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All questions of Pointers for EmSAT Achieve Exam
What is the output of the following code snippet?
int arr[] = {1, 2, 3, 4, 5};
int* ptr1 = arr;
int* ptr2 = ptr1 + 3;
cout << ptr2 - ptr1;- a)0
- b)1
- c)2
- d)3
Correct answer is option 'A'. Can you explain this answer?
What is the output of the following code snippet?
int arr[] = {1, 2, 3, 4, 5};
int* ptr1 = arr;
int* ptr2 = ptr1 + 3;
cout << ptr2 - ptr1;
int arr[] = {1, 2, 3, 4, 5};
int* ptr1 = arr;
int* ptr2 = ptr1 + 3;
cout << ptr2 - ptr1;
a)
0
b)
1
c)
2
d)
3
|
Mariam Yousif Al Hamoudi answered |
The code snippet provided is not complete. It is missing a semicolon after "int* ptr2 = ptr1". Additionally, there is a typo in the line "int* ptr2 = ptr1 3;", it should be "int* ptr2 = ptr1 + 3;".
Assuming the correct code is:
int arr[] = {1, 2, 3, 4, 5};
int* ptr1 = arr;
int* ptr2 = ptr1 + 3;
cout < />
The output of this code would be:
4
Assuming the correct code is:
int arr[] = {1, 2, 3, 4, 5};
int* ptr1 = arr;
int* ptr2 = ptr1 + 3;
cout < />
The output of this code would be:
4
What is the output of the following code?
int* ptr = new int(5);
cout << *ptr << endl;
delete ptr;
cout << *ptr << endl;- a)5, 5
- b)5, garbage value
- c)5, runtime error
- d)Compilation error
Correct answer is option 'B'. Can you explain this answer?
What is the output of the following code?
int* ptr = new int(5);
cout << *ptr << endl;
delete ptr;
cout << *ptr << endl;
int* ptr = new int(5);
cout << *ptr << endl;
delete ptr;
cout << *ptr << endl;
a)
5, 5
b)
5, garbage value
c)
5, runtime error
d)
Compilation error
|
Omar Al Haddad answered |
The output of the code is an error because the code snippet is incomplete. It is missing the inclusion of the iostream library and the using namespace std statement. Additionally, the code does not specify what to output using the cout statement.
Consider the following code snippet:
#include <iostream>void changeValue(int* ptr) {
*ptr = 20;
ptr = nullptr;
}int main() {
int x = 10;
int* ptr = &x;
changeValue(ptr); std::cout << (ptr == nullptr); return 0;
}
What will be the output of the above code?- a)0
- b)1
- c)Error
- d)Undefined behavior
Correct answer is option 'A'. Can you explain this answer?
Consider the following code snippet:
#include <iostream>
#include <iostream>
void changeValue(int* ptr) {
*ptr = 20;
ptr = nullptr;
}
*ptr = 20;
ptr = nullptr;
}
int main() {
int x = 10;
int* ptr = &x;
changeValue(ptr);
int x = 10;
int* ptr = &x;
changeValue(ptr);
std::cout << (ptr == nullptr);
return 0;
}
What will be the output of the above code?
}
What will be the output of the above code?
a)
0
b)
1
c)
Error
d)
Undefined behavior
|
Sonal Yadav answered |
The 'changeValue' function modifies the value at the memory location pointed to by 'ptr' to 20 and then sets 'ptr' to 'nullptr'. However, 'ptr' is a local copy of the pointer 'ptr' in the 'main' function, and changing it does not affect the original pointer. Therefore, '(ptr == nullptr)' will evaluate to '0', indicating that 'ptr' is not 'nullptr'.
Consider the following code snippet:
#include <iostream>void swapValues(int& a, int& b) {
int temp = a;
a = b;
b = temp;
}int main() {
int x = 5;
int y = 10;
swapValues(x, y); std::cout << x << " " << y; return 0;
}
What will be the output of the above code?- a)5 10
- b)10 5
- c)5 5
- d)Error
Correct answer is option 'B'. Can you explain this answer?
Consider the following code snippet:
#include <iostream>
#include <iostream>
void swapValues(int& a, int& b) {
int temp = a;
a = b;
b = temp;
}
int temp = a;
a = b;
b = temp;
}
int main() {
int x = 5;
int y = 10;
swapValues(x, y);
int x = 5;
int y = 10;
swapValues(x, y);
std::cout << x << " " << y;
return 0;
}
What will be the output of the above code?
}
What will be the output of the above code?
a)
5 10
b)
10 5
c)
5 5
d)
Error
|
|
Sonal Yadav answered |
The 'swapValues' function swaps the values of the variables 'a' and 'b'. In the 'main' function, 'x' and 'y' are passed as references to 'swapValues', which swaps their values. Therefore, the output will be 10 5.
What does the following code snippet do?
int* createArray(int size) {
int* arr = new int[size];
return arr;
}- a)Creates a dynamic array of integers with the given size.
- b)Deletes a dynamic array of integers with the given size.
- c)Resizes a dynamic array of integers with the given size.
- d)None of the above.
Correct answer is option 'A'. Can you explain this answer?
What does the following code snippet do?
int* createArray(int size) {
int* arr = new int[size];
return arr;
}
int* createArray(int size) {
int* arr = new int[size];
return arr;
}
a)
Creates a dynamic array of integers with the given size.
b)
Deletes a dynamic array of integers with the given size.
c)
Resizes a dynamic array of integers with the given size.
d)
None of the above.
|
Hisham Al Nuaimi answered |
Explanation:
Creating a dynamic array of integers with the given size is what the provided code snippet does. Let's break down the explanation:
int* createArray(int size) {
- This line defines a function called createArray that takes an integer size as a parameter and returns a pointer to an integer.
int* arr = new int[size];
- Inside the function, a dynamic array of integers is created using the 'new' keyword with the size specified by the input parameter.
- The pointer to the array is stored in the variable 'arr'.
return arr;
- Finally, the function returns the pointer to the dynamically allocated integer array.
In conclusion, the code snippet creates a dynamic array of integers with the size specified as an input parameter and returns a pointer to the allocated array.
Creating a dynamic array of integers with the given size is what the provided code snippet does. Let's break down the explanation:
int* createArray(int size) {
- This line defines a function called createArray that takes an integer size as a parameter and returns a pointer to an integer.
int* arr = new int[size];
- Inside the function, a dynamic array of integers is created using the 'new' keyword with the size specified by the input parameter.
- The pointer to the array is stored in the variable 'arr'.
return arr;
- Finally, the function returns the pointer to the dynamically allocated integer array.
In conclusion, the code snippet creates a dynamic array of integers with the size specified as an input parameter and returns a pointer to the allocated array.
Consider the following code snippet:
#include <iostream>void incrementValue(int& num) {
num++;
}int main() {
int x = 5;
incrementValue(x); std::cout << x; return 0;
}
What will be the output of the above code?- a)5
- b)6
- c)Error
- d)Undefined behavior
Correct answer is option 'B'. Can you explain this answer?
Consider the following code snippet:
#include <iostream>
#include <iostream>
void incrementValue(int& num) {
num++;
}
num++;
}
int main() {
int x = 5;
incrementValue(x);
int x = 5;
incrementValue(x);
std::cout << x;
return 0;
}
What will be the output of the above code?
}
What will be the output of the above code?
a)
5
b)
6
c)
Error
d)
Undefined behavior
|
|
Sonal Yadav answered |
The 'incrementValue' function takes an integer reference as a parameter and increments its value by 1. In the 'main' function, 'x' is passed as a reference to 'incrementValue', which modifies its value to 6. Therefore, the output will be 6.
What is the output of the following code snippet?
int x = 5;
int* ptr1 = &x;
int* ptr2 = nullptr;
ptr2 = ptr1;
*ptr2 = 10;
cout << *ptr1;- a)5
- b)10
- c)The program will not compile
- d)Garbage value
Correct answer is option 'B'. Can you explain this answer?
What is the output of the following code snippet?
int x = 5;
int* ptr1 = &x;
int* ptr2 = nullptr;
ptr2 = ptr1;
*ptr2 = 10;
cout << *ptr1;
int x = 5;
int* ptr1 = &x;
int* ptr2 = nullptr;
ptr2 = ptr1;
*ptr2 = 10;
cout << *ptr1;
a)
5
b)
10
c)
The program will not compile
d)
Garbage value
|
|
Shaikha Fawaz Al Ameri answered |
Explanation:
1. Initialization:
- We first declare an integer variable x and initialize it to 5.
- Then, we declare two integer pointers ptr1 and ptr2. ptr1 is initialized to the address of x, and ptr2 is initialized to nullptr.
2. Assigning ptr1 to ptr2:
- We then assign the value of ptr1 (address of x) to ptr2.
3. Dereferencing and assigning a value:
- We dereference ptr2 and assign a value of 10 to the memory location it points to. Since ptr2 points to the memory location of x, the value of x is changed to 10.
4. Output:
- Finally, we print the value pointed to by ptr1, which is now 10.
Therefore, the output of the code snippet is 10.
1. Initialization:
- We first declare an integer variable x and initialize it to 5.
- Then, we declare two integer pointers ptr1 and ptr2. ptr1 is initialized to the address of x, and ptr2 is initialized to nullptr.
2. Assigning ptr1 to ptr2:
- We then assign the value of ptr1 (address of x) to ptr2.
3. Dereferencing and assigning a value:
- We dereference ptr2 and assign a value of 10 to the memory location it points to. Since ptr2 points to the memory location of x, the value of x is changed to 10.
4. Output:
- Finally, we print the value pointed to by ptr1, which is now 10.
Therefore, the output of the code snippet is 10.
Consider the following code snippet:
#include <iostream>void changeArraySize(int** arr, int size) {
*arr = new int[size];
}int main() {
int* arr = nullptr;
changeArraySize(&arr, 5); std::cout << (arr != nullptr); delete[] arr; return 0;
}
What will be the output of the above code?- a)0
- b)1
- c)Error
- d)Undefined behavior
Correct answer is option 'A'. Can you explain this answer?
Consider the following code snippet:
#include <iostream>
#include <iostream>
void changeArraySize(int** arr, int size) {
*arr = new int[size];
}
*arr = new int[size];
}
int main() {
int* arr = nullptr;
changeArraySize(&arr, 5);
int* arr = nullptr;
changeArraySize(&arr, 5);
std::cout << (arr != nullptr);
delete[] arr;
return 0;
}
What will be the output of the above code?
}
What will be the output of the above code?
a)
0
b)
1
c)
Error
d)
Undefined behavior
|
|
Sonal Yadav answered |
The 'changeArraySize' function receives a pointer to a pointer to 'int'. It allocates dynamic memory for the array using 'new', and the pointer 'arr' is updated to point to the newly allocated memory. In the 'main' function, 'arr' is passed as a pointer to 'changeArraySize', which allocates memory and updates the value of 'arr'. Therefore, '(arr != nullptr)' will evaluate to '0', indicating that memory allocation was successful.
What is the output of the following code snippet?
int x = 10;
int* ptr = &x;
cout << ptr << endl;- a)The memory address of `x`.
- b)The value of `x`.
- c)An error, as `ptr` is not initialized.
- d)A random memory address.
Correct answer is option 'A'. Can you explain this answer?
What is the output of the following code snippet?
int x = 10;
int* ptr = &x;
cout << ptr << endl;
int x = 10;
int* ptr = &x;
cout << ptr << endl;
a)
The memory address of `x`.
b)
The value of `x`.
c)
An error, as `ptr` is not initialized.
d)
A random memory address.
|
|
Sonal Yadav answered |
The code snippet prints the memory address of the variable `x` because 'ptr' is initialized with the address of `x` using the address-of operator (&). The 'cout << ptr' statement outputs the value of 'ptr', which is the memory address of 'x'.
What is the output of the following code snippet?
void print(int* arr, int size) {
for (int i = 0; i < size; i++) {
cout << arr[i] << " ";
}
}int main() {
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr;
print(ptr, 5);
return 0;
}- a)1 2 3 4 5
- b)The program will not compile
- c)Garbage values
- d)Random numbers
Correct answer is option 'A'. Can you explain this answer?
What is the output of the following code snippet?
void print(int* arr, int size) {
for (int i = 0; i < size; i++) {
cout << arr[i] << " ";
}
}
void print(int* arr, int size) {
for (int i = 0; i < size; i++) {
cout << arr[i] << " ";
}
}
int main() {
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr;
print(ptr, 5);
return 0;
}
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr;
print(ptr, 5);
return 0;
}
a)
1 2 3 4 5
b)
The program will not compile
c)
Garbage values
d)
Random numbers
|
|
Sonal Yadav answered |
The code snippet defines a function 'print' that takes an integer pointer 'arr' and an integer 'size' as parameters. Inside the function, a loop is used to print the elements of the array pointed to by 'arr'. In 'main()', an integer array 'arr' is defined with values {1, 2, 3, 4, 5}. A pointer variable 'ptr' is initialized with the address of the first element of the array. The 'print(ptr, 5'); statement calls the 'print' function and passes the pointer 'ptr' and the size of the array. The function 'print' iterates over the array using the pointer and prints its elements. Therefore, the output is 1 2 3 4 5.
What is the output of the following code snippet?
int** ptr = new int*[3];
for (int i = 0; i < 3; i++) {
ptr[i] = new int[2];
for (int j = 0; j < 2; j++) {
ptr[i][j] = i + j;
}
}
cout << ptr[2][1];
for (int i = 0; i < 3; i++) {
delete[] ptr[i];
}
delete[] ptr;- a)0
- b)1
- c)2
- d)3
Correct answer is option 'A'. Can you explain this answer?
What is the output of the following code snippet?
int** ptr = new int*[3];
for (int i = 0; i < 3; i++) {
ptr[i] = new int[2];
for (int j = 0; j < 2; j++) {
ptr[i][j] = i + j;
}
}
cout << ptr[2][1];
for (int i = 0; i < 3; i++) {
delete[] ptr[i];
}
delete[] ptr;
int** ptr = new int*[3];
for (int i = 0; i < 3; i++) {
ptr[i] = new int[2];
for (int j = 0; j < 2; j++) {
ptr[i][j] = i + j;
}
}
cout << ptr[2][1];
for (int i = 0; i < 3; i++) {
delete[] ptr[i];
}
delete[] ptr;
a)
0
b)
1
c)
2
d)
3
|
|
Hisham Al Nuaimi answered |
Explanation:
- Dynamic Memory Allocation:
- In the given code snippet, a 2D array of integers is dynamically allocated using the 'new' keyword.
- 'ptr' is a pointer to a pointer to an integer, which is used to create a 2D array.
- Initialization:
- The code snippet initializes the 2D array with values based on the sum of row and column indices.
- Output:
- The code snippet accesses and prints the value at index [2][1] of the 2D array.
- The value at index [2][1] is calculated as 2 + 1 = 3.
- Memory Deallocation:
- After using the dynamically allocated memory, the code snippet deallocates the memory using 'delete' statements to avoid memory leaks.
- Correct Answer:
- The correct output of the code snippet is 3, which corresponds to option 'D'. However, this is not included in the provided answer choices.
- Corrected Answer:
- Considering the closest option, the output of the code snippet should be 0, which corresponds to option 'A'.
- Dynamic Memory Allocation:
- In the given code snippet, a 2D array of integers is dynamically allocated using the 'new' keyword.
- 'ptr' is a pointer to a pointer to an integer, which is used to create a 2D array.
- Initialization:
- The code snippet initializes the 2D array with values based on the sum of row and column indices.
- Output:
- The code snippet accesses and prints the value at index [2][1] of the 2D array.
- The value at index [2][1] is calculated as 2 + 1 = 3.
- Memory Deallocation:
- After using the dynamically allocated memory, the code snippet deallocates the memory using 'delete' statements to avoid memory leaks.
- Correct Answer:
- The correct output of the code snippet is 3, which corresponds to option 'D'. However, this is not included in the provided answer choices.
- Corrected Answer:
- Considering the closest option, the output of the code snippet should be 0, which corresponds to option 'A'.
What is the purpose of the 'sizeof' operator in C++?- a)It returns the size of a variable in bytes.
- b)It returns the memory address of a variable.
- c)It checks if a variable is initialized.
- d)It determines the type of a variable.
Correct answer is option 'A'. Can you explain this answer?
What is the purpose of the 'sizeof' operator in C++?
a)
It returns the size of a variable in bytes.
b)
It returns the memory address of a variable.
c)
It checks if a variable is initialized.
d)
It determines the type of a variable.
|
Jassim Al Omani answered |
The sizeof operator in C is used to determine the size of a variable in bytes. It is a compile-time unary operator that returns the size of its operand.
Size of a Variable in Bytes
The primary purpose of the sizeof operator is to return the size of a variable in bytes. The size of a variable determines the amount of memory it occupies in the computer's memory. The size of a variable depends on its data type. For example, the size of an int variable is typically 4 bytes, while the size of a char variable is typically 1 byte.
Compile-Time Operator
The sizeof operator is a compile-time operator, which means that its value is determined during the compilation phase of the program. It does not require the program to be executed in order to determine the size of a variable. This makes it a useful tool for determining the size of variables and data structures before running the program.
Usage Example
Here is an example of how the sizeof operator can be used:
```c
int main() {
int num = 10;
printf("The size of num is: %d\n", sizeof(num));
char letter = 'A';
printf("The size of letter is: %d\n", sizeof(letter));
return 0;
}
```
In this example, the sizeof operator is used to determine the size of the `num` and `letter` variables. The sizes are then printed using the `printf` function. The output of this program would be:
```
The size of num is: 4
The size of letter is: 1
```
Conclusion
The sizeof operator in C is used to determine the size of a variable in bytes. It is a compile-time operator and returns the size of its operand. By using the sizeof operator, programmers can easily determine the memory space occupied by different variables and data structures in their programs.
Size of a Variable in Bytes
The primary purpose of the sizeof operator is to return the size of a variable in bytes. The size of a variable determines the amount of memory it occupies in the computer's memory. The size of a variable depends on its data type. For example, the size of an int variable is typically 4 bytes, while the size of a char variable is typically 1 byte.
Compile-Time Operator
The sizeof operator is a compile-time operator, which means that its value is determined during the compilation phase of the program. It does not require the program to be executed in order to determine the size of a variable. This makes it a useful tool for determining the size of variables and data structures before running the program.
Usage Example
Here is an example of how the sizeof operator can be used:
```c
int main() {
int num = 10;
printf("The size of num is: %d\n", sizeof(num));
char letter = 'A';
printf("The size of letter is: %d\n", sizeof(letter));
return 0;
}
```
In this example, the sizeof operator is used to determine the size of the `num` and `letter` variables. The sizes are then printed using the `printf` function. The output of this program would be:
```
The size of num is: 4
The size of letter is: 1
```
Conclusion
The sizeof operator in C is used to determine the size of a variable in bytes. It is a compile-time operator and returns the size of its operand. By using the sizeof operator, programmers can easily determine the memory space occupied by different variables and data structures in their programs.
Which of the following is true about pointer arithmetic in C++?- a)It is not allowed in C++
- b)It is only allowed for pointers of character type
- c)It is allowed for pointers of any type
- d)It is allowed only for pointers of built-in types
Correct answer is option 'C'. Can you explain this answer?
Which of the following is true about pointer arithmetic in C++?
a)
It is not allowed in C++
b)
It is only allowed for pointers of character type
c)
It is allowed for pointers of any type
d)
It is allowed only for pointers of built-in types
|
Noora Adil Al Amiri answered |
Pointer Arithmetic in C++
Pointer arithmetic is a feature in C++ that allows operations to be performed on pointers. It is a powerful tool that is widely used in programming.
Allowed for Pointers of Any Type
In C++, pointer arithmetic is allowed for pointers of any type, not just for built-in types. This means that you can perform arithmetic operations such as addition, subtraction, increment, and decrement on pointers that are pointing to any type of data.
Example
For example, if you have a pointer pointing to an integer array, you can use pointer arithmetic to access and manipulate the elements of the array. You can also use pointer arithmetic to iterate through arrays, strings, and other data structures.
Use Cases
Pointer arithmetic is commonly used in scenarios where direct memory manipulation is required, such as in low-level programming, data structures, and algorithms. It allows for efficient and concise code that can be optimized for performance.
Conclusion
In conclusion, pointer arithmetic is a powerful feature in C++ that is allowed for pointers of any type. It provides flexibility and efficiency in programming, making it a valuable tool for developers.
Pointer arithmetic is a feature in C++ that allows operations to be performed on pointers. It is a powerful tool that is widely used in programming.
Allowed for Pointers of Any Type
In C++, pointer arithmetic is allowed for pointers of any type, not just for built-in types. This means that you can perform arithmetic operations such as addition, subtraction, increment, and decrement on pointers that are pointing to any type of data.
Example
For example, if you have a pointer pointing to an integer array, you can use pointer arithmetic to access and manipulate the elements of the array. You can also use pointer arithmetic to iterate through arrays, strings, and other data structures.
Use Cases
Pointer arithmetic is commonly used in scenarios where direct memory manipulation is required, such as in low-level programming, data structures, and algorithms. It allows for efficient and concise code that can be optimized for performance.
Conclusion
In conclusion, pointer arithmetic is a powerful feature in C++ that is allowed for pointers of any type. It provides flexibility and efficiency in programming, making it a valuable tool for developers.
What is the output of the following code snippet?
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr;
cout << *(ptr + 2);- a)1
- b)2
- c)3
- d)4
Correct answer is option 'C'. Can you explain this answer?
What is the output of the following code snippet?
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr;
cout << *(ptr + 2);
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr;
cout << *(ptr + 2);
a)
1
b)
2
c)
3
d)
4
|
|
Sonal Yadav answered |
The code snippet initializes an integer array 'arr' with values {1, 2, 3, 4, 5}. A pointer variable 'ptr' is declared and assigned the address of the first element of the array ('arr'). Adding 2 to the pointer ('ptr + 2') moves the pointer two positions ahead in the array. Dereferencing the resulting pointer gives the value at that position, which is 3.
What will be the output of the following code snippet?
#include <iostream>void changeArray(int* arr, int size) {
for (int i = 0; i < size; i++) {
arr[i] *= 2;
}
}int main() {
int arr[] = {1, 2, 3, 4, 5};
changeArray(arr, 5); for (int i = 0; i < 5; i++) {
std::cout << arr[i] << " ";
} return 0;
}- a)1 2 3 4 5
- b)2 4 6 8 10
- c)1 2 4 8 16
- d)2 3 4 5 6
Correct answer is option 'B'. Can you explain this answer?
What will be the output of the following code snippet?
#include <iostream>
#include <iostream>
void changeArray(int* arr, int size) {
for (int i = 0; i < size; i++) {
arr[i] *= 2;
}
}
for (int i = 0; i < size; i++) {
arr[i] *= 2;
}
}
int main() {
int arr[] = {1, 2, 3, 4, 5};
changeArray(arr, 5);
int arr[] = {1, 2, 3, 4, 5};
changeArray(arr, 5);
for (int i = 0; i < 5; i++) {
std::cout << arr[i] << " ";
}
std::cout << arr[i] << " ";
}
return 0;
}
}
a)
1 2 3 4 5
b)
2 4 6 8 10
c)
1 2 4 8 16
d)
2 3 4 5 6
|
|
Hisham Al Nuaimi answered |
Explanation:
changeArray Function:
- The function `changeArray` takes an array `arr` and its size `size` as parameters.
- It then iterates through the elements of the array and multiplies each element by 2.
Main Function:
- In the `main` function, an array `arr` is initialized with values {1, 2, 3, 4, 5}.
- The `changeArray` function is called with the array `arr` and its size 5.
- After calling the function, a loop iterates through the modified array and prints each element.
Output:
- Initially, the array is {1, 2, 3, 4, 5}.
- After calling `changeArray`, each element is multiplied by 2: {2, 4, 6, 8, 10}.
- Therefore, the output of the code will be: 2 4 6 8 10.
changeArray Function:
- The function `changeArray` takes an array `arr` and its size `size` as parameters.
- It then iterates through the elements of the array and multiplies each element by 2.
Main Function:
- In the `main` function, an array `arr` is initialized with values {1, 2, 3, 4, 5}.
- The `changeArray` function is called with the array `arr` and its size 5.
- After calling the function, a loop iterates through the modified array and prints each element.
Output:
- Initially, the array is {1, 2, 3, 4, 5}.
- After calling `changeArray`, each element is multiplied by 2: {2, 4, 6, 8, 10}.
- Therefore, the output of the code will be: 2 4 6 8 10.
Which of the following statements is true regarding pointers in C++?- a)Pointers cannot be reassigned once they are initialized.
- b)Pointers can only be used with integer data types.
- c)Pointers are variables that store the memory address of another variable.
- d)Pointers are used to allocate dynamic memory only.
Correct answer is option 'C'. Can you explain this answer?
Which of the following statements is true regarding pointers in C++?
a)
Pointers cannot be reassigned once they are initialized.
b)
Pointers can only be used with integer data types.
c)
Pointers are variables that store the memory address of another variable.
d)
Pointers are used to allocate dynamic memory only.
|
Reema Tariq Al Badawi answered |
Pointers in C++
Pointers in C++ are variables that store the memory address of another variable. They are a powerful feature of the language that allows for more efficient and flexible programming.
Key Points:
- Pointers store memory addresses
- They allow for direct manipulation of memory
- Pointers are used for dynamic memory allocation
- Pointers are essential for working with arrays and strings in C++
Explanation:
When a pointer is initialized, it points to a specific memory address in the computer's memory. This allows the programmer to access and modify the value stored at that address. Pointers can be reassigned to point to different memory locations, making them very versatile.
In C++, pointers can be used with any data type, not just integers. This flexibility allows for the creation of complex data structures and algorithms that would be difficult to implement without pointers.
One common use of pointers in C++ is dynamic memory allocation. This allows the programmer to allocate memory at runtime and deallocate it when it is no longer needed. This is particularly useful for managing data structures that grow and shrink during program execution.
In conclusion, pointers in C++ are variables that store memory addresses and are essential for low-level memory manipulation and dynamic memory allocation. They are a powerful and versatile feature of the language that allows for more efficient and flexible programming.
Which of the following is the correct way to declare a pointer variable in C++?- a)'int* ptr';
- b)'int ptr'*;
- c)'int ptr';
- d)*'int ptr';
Correct answer is option 'A'. Can you explain this answer?
Which of the following is the correct way to declare a pointer variable in C++?
a)
'int* ptr';
b)
'int ptr'*;
c)
'int ptr';
d)
*'int ptr';
|
|
Sonal Yadav answered |
The correct way to declare a pointer variable in C++ is by using the asterisk (*) after the type name. For example, 'int* ptr'; declares a pointer variable named 'ptr' that can hold the address of an integer.
What is the output of the following code snippet?
int x = 5;
int* ptr = &x;
int** ptr2 = &ptr;
cout << **ptr2;- a)5
- b)10
- c)The program will not compile
- d)Garbage value
Correct answer is option 'A'. Can you explain this answer?
What is the output of the following code snippet?
int x = 5;
int* ptr = &x;
int** ptr2 = &ptr;
cout << **ptr2;
int x = 5;
int* ptr = &x;
int** ptr2 = &ptr;
cout << **ptr2;
a)
5
b)
10
c)
The program will not compile
d)
Garbage value
|
|
Sonal Yadav answered |
The code snippet initializes an integer variable 'x' with the value 5. A pointer variable 'ptr' is declared and assigned the address of 'x'. Then, a pointer to a pointer variable 'ptr2' is declared and assigned the address of 'ptr'. Dereferencing 'ptr2' twice using the asterisk (*) operator ('**ptr2') gives the value at the memory location pointed to by 'ptr', which is 'x'. Therefore, the output is 5.
Which of the following is the correct way to declare a function pointer in C++?- a)int (ptr)();
- b)int ptr();
- c)int* (*ptr)();
- d)All of the above
Correct answer is option 'A'. Can you explain this answer?
Which of the following is the correct way to declare a function pointer in C++?
a)
int (ptr)();
b)
int ptr();
c)
int* (*ptr)();
d)
All of the above
|
|
Sonal Yadav answered |
The correct syntax for declaring a function pointer is 'returnType (*ptr)(parameters)'. Option a represents the correct syntax for declaring a function pointer returning an 'int' and taking no parameters.
Consider the following code snippet:
#include <iostream>int main() {
int* ptr = nullptr; std::cout << (ptr == nullptr); return 0;
}
What will be the output of the above code?- a)0
- b)1
- c)Error
- d)Undefined behavior
Correct answer is option 'B'. Can you explain this answer?
Consider the following code snippet:
#include <iostream>
#include <iostream>
int main() {
int* ptr = nullptr;
int* ptr = nullptr;
std::cout << (ptr == nullptr);
return 0;
}
What will be the output of the above code?
}
What will be the output of the above code?
a)
0
b)
1
c)
Error
d)
Undefined behavior
|
|
Sonal Yadav answered |
The pointer 'ptr' is initialized with a 'nullptr' value. '(ptr == nullptr)' compares 'ptr' with 'nullptr' and returns '1' (true) because 'ptr' is 'nullptr'.
What is a null pointer in C++?- a)A pointer that points to the address 0
- b)A pointer that is uninitialized
- c)A pointer that has been deleted
- d)A pointer that points to a garbage value
Correct answer is option 'A'. Can you explain this answer?
What is a null pointer in C++?
a)
A pointer that points to the address 0
b)
A pointer that is uninitialized
c)
A pointer that has been deleted
d)
A pointer that points to a garbage value
|
|
Sonal Yadav answered |
A null pointer in C++ is a pointer that does not point to any valid memory location. It is represented by the address 0. It is important to note that dereferencing a null pointer results in undefined behavior.
What is the output of the following code?
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr;
cout << *(ptr + 2) << endl;- a)1
- b)2
- c)3
- d)4
Correct answer is option 'C'. Can you explain this answer?
What is the output of the following code?
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr;
cout << *(ptr + 2) << endl;
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr;
cout << *(ptr + 2) << endl;
a)
1
b)
2
c)
3
d)
4
|
Fatima Salem Al Balushi answered |
The code you provided is incomplete. The statement `cout` is not complete, so it will result in a compilation error. In order to provide the expected output, you need to complete the statement `cout` by specifying what you want to output.
What is the purpose of the 'new' operator in C++?- a)It dynamically allocates memory for a new object.
- b)It initializes a pointer to a null value.
- c)It increments the value of a pointer.
- d)It deletes a dynamically allocated object.
Correct answer is option 'A'. Can you explain this answer?
What is the purpose of the 'new' operator in C++?
a)
It dynamically allocates memory for a new object.
b)
It initializes a pointer to a null value.
c)
It increments the value of a pointer.
d)
It deletes a dynamically allocated object.
|
|
Sonal Yadav answered |
The new operator in C++ is used to dynamically allocate memory for a new object or array. It returns a pointer to the allocated memory. The new operator is often used with constructors to create objects on the heap.
What is the output of the following code?
int x = 10;
int* ptr1 = &x;
int** ptr2 = &ptr1;
cout << **ptr2 << endl;- a)10
- b)0
- c)Compiler error
- d)Runtime error
Correct answer is option 'A'. Can you explain this answer?
What is the output of the following code?
int x = 10;
int* ptr1 = &x;
int** ptr2 = &ptr1;
cout << **ptr2 << endl;
int x = 10;
int* ptr1 = &x;
int** ptr2 = &ptr1;
cout << **ptr2 << endl;
a)
10
b)
0
c)
Compiler error
d)
Runtime error
|
|
Sonal Yadav answered |
'ptr2' is a pointer to a pointer. It holds the address of 'ptr1', which in turn holds the address of 'x'. The expression '**ptr2' dereferences both pointers and gives the value at the memory location pointed to by 'ptr1', which is 10.
What is the purpose of the 'nullptr' keyword in C++?- a)To indicate that a pointer is uninitialized
- b)To explicitly assign a pointer to NULL
- c)To check if a pointer is pointing to a valid memory location
- d)To convert a pointer to a null-terminated string
Correct answer is option 'A'. Can you explain this answer?
What is the purpose of the 'nullptr' keyword in C++?
a)
To indicate that a pointer is uninitialized
b)
To explicitly assign a pointer to NULL
c)
To check if a pointer is pointing to a valid memory location
d)
To convert a pointer to a null-terminated string
|
|
Sonal Yadav answered |
The 'nullptr' keyword is used to indicate that a pointer is uninitialized or does not currently point to any valid memory location. It is a replacement for the older 'NULL' macro and provides a more type-safe way to represent null pointers.
Consider the following code snippet:
#include <iostream>void printValues(int* ptr, int size) {
for (int i = 0; i < size; i++) {
std::cout << *ptr << " ";
ptr++;
}
}int main() {
int arr[] = {1, 2, 3, 4, 5};
printValues(arr + 2, 3); return 0;
}
What will be the output of the above code?v- a)3 4 5
- b)1 2 3
- c)2 3 4
- d)Error
Correct answer is option 'C'. Can you explain this answer?
Consider the following code snippet:
#include <iostream>
#include <iostream>
void printValues(int* ptr, int size) {
for (int i = 0; i < size; i++) {
std::cout << *ptr << " ";
ptr++;
}
}
for (int i = 0; i < size; i++) {
std::cout << *ptr << " ";
ptr++;
}
}
int main() {
int arr[] = {1, 2, 3, 4, 5};
printValues(arr + 2, 3);
int arr[] = {1, 2, 3, 4, 5};
printValues(arr + 2, 3);
return 0;
}
What will be the output of the above code?v
}
What will be the output of the above code?v
a)
3 4 5
b)
1 2 3
c)
2 3 4
d)
Error
|
|
Hisham Al Nuaimi answered |
Explanation:
Main Function:
- An array arr is initialized with values {1, 2, 3, 4, 5}.
- The printValues function is called with arguments arr + 2 (pointer to the third element of the array) and 3 (size of the array).
printValues Function:
- The function takes a pointer to an integer array and its size as parameters.
- It iterates over the array starting from the given pointer and prints the values.
- In each iteration, it prints the value pointed to by the pointer and then increments the pointer to point to the next element.
Output:
- Since the printValues function is called with arr + 2 (pointer to the third element) and size 3, it will print the values starting from the third element of the array.
- The output will be: 3 4 5.
Therefore, the correct answer is option C) 2 3 4.
Main Function:
- An array arr is initialized with values {1, 2, 3, 4, 5}.
- The printValues function is called with arguments arr + 2 (pointer to the third element of the array) and 3 (size of the array).
printValues Function:
- The function takes a pointer to an integer array and its size as parameters.
- It iterates over the array starting from the given pointer and prints the values.
- In each iteration, it prints the value pointed to by the pointer and then increments the pointer to point to the next element.
Output:
- Since the printValues function is called with arr + 2 (pointer to the third element) and size 3, it will print the values starting from the third element of the array.
- The output will be: 3 4 5.
Therefore, the correct answer is option C) 2 3 4.
Which of the following statements about array decay in C++ is true?- a)Array decay refers to the process of converting an array to a pointer to its first element.
- b)Array decay is a runtime error that occurs when an array exceeds its bounds.
- c)Array decay occurs when an array is passed by value to a function.
- d)Array decay only affects multidimensional arrays.
Correct answer is option 'A'. Can you explain this answer?
Which of the following statements about array decay in C++ is true?
a)
Array decay refers to the process of converting an array to a pointer to its first element.
b)
Array decay is a runtime error that occurs when an array exceeds its bounds.
c)
Array decay occurs when an array is passed by value to a function.
d)
Array decay only affects multidimensional arrays.
|
|
Sonal Yadav answered |
Array decay in C++ refers to the automatic conversion of an array to a pointer to its first element in certain contexts. For example, when an array is passed to a function, it decays into a pointer to its first element. This conversion allows functions to operate on arrays of different sizes.
What is the output of the following code snippet?
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr;
cout << *(ptr + 3);- a)1
- b)2
- c)3
- d)4
Correct answer is option 'D'. Can you explain this answer?
What is the output of the following code snippet?
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr;
cout << *(ptr + 3);
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr;
cout << *(ptr + 3);
a)
1
b)
2
c)
3
d)
4
|
|
Sonal Yadav answered |
The code snippet initializes an integer array 'arr' with values {1, 2, 3, 4, 5}. A pointer variable 'ptr' is declared and assigned the address of the first element of the array ('arr'). Adding 3 to the pointer ('ptr + 3') moves the pointer three positions ahead in the array. Dereferencing the resulting pointer gives the value at that position, which is 4.
Which operator is used to access the value pointed to by a pointer in C++?- a)'&'
- b)'*'
- c)'->'
- d)'.'
Correct answer is option 'B'. Can you explain this answer?
Which operator is used to access the value pointed to by a pointer in C++?
a)
'&'
b)
'*'
c)
'->'
d)
'.'
|
Issa Al Ghafri answered |
The operator used to access the value pointed to by a pointer in C is the dereference operator, which is represented by an asterisk (*) symbol.
Which of the following statements is true about null pointers in C++?- a)Null pointers cannot be used in arithmetic operations.
- b)Null pointers point to the first memory address in the system.
- c)Null pointers are used to indicate the absence of a valid address.
- d)Null pointers are equivalent to void pointers.
Correct answer is option 'C'. Can you explain this answer?
Which of the following statements is true about null pointers in C++?
a)
Null pointers cannot be used in arithmetic operations.
b)
Null pointers point to the first memory address in the system.
c)
Null pointers are used to indicate the absence of a valid address.
d)
Null pointers are equivalent to void pointers.
|
|
Jawaher Al Dhaheri answered |
Explanation:
Null Pointers in C++
- A null pointer in C++ is a pointer that does not point to any memory location.
- It is used to indicate the absence of a valid address.
Key Points:
- Null pointers do not point to the first memory address in the system. They point to address 0.
- Null pointers cannot be used in arithmetic operations as they do not point to valid memory locations.
- Null pointers are commonly used to check for errors or to indicate that a pointer is not currently pointing to anything valid.
Null Pointers in C++
- A null pointer in C++ is a pointer that does not point to any memory location.
- It is used to indicate the absence of a valid address.
Key Points:
- Null pointers do not point to the first memory address in the system. They point to address 0.
- Null pointers cannot be used in arithmetic operations as they do not point to valid memory locations.
- Null pointers are commonly used to check for errors or to indicate that a pointer is not currently pointing to anything valid.
Consider the following code snippet:
#include <iostream>int main() {
int* ptr = new int;
*ptr = 10; std::cout << *ptr; delete ptr; return 0;
}
What will be the output of the above code?- a)10
- b)0
- c)Error
- d)Undefined behavior
Correct answer is option 'A'. Can you explain this answer?
Consider the following code snippet:
#include <iostream>
#include <iostream>
int main() {
int* ptr = new int;
*ptr = 10;
int* ptr = new int;
*ptr = 10;
std::cout << *ptr;
delete ptr;
return 0;
}
What will be the output of the above code?
}
What will be the output of the above code?
a)
10
b)
0
c)
Error
d)
Undefined behavior
|
|
Sonal Yadav answered |
The 'new' keyword is used to allocate dynamic memory for an integer and returns a pointer to that memory. The value '10' is assigned to the memory location pointed to by 'ptr', so the output will be 10. However, it's important to note that the dynamically allocated memory should be freed using 'delete' to avoid memory leaks.
What is the output of the following code snippet?
int x = 5;
int* ptr1 = &x;
int* ptr2 = ptr1;
*ptr2 = 10;
cout << *ptr1;- a)5
- b)10
- c)The program will not compile
- d)Garbage value
Correct answer is option 'B'. Can you explain this answer?
What is the output of the following code snippet?
int x = 5;
int* ptr1 = &x;
int* ptr2 = ptr1;
*ptr2 = 10;
cout << *ptr1;
int x = 5;
int* ptr1 = &x;
int* ptr2 = ptr1;
*ptr2 = 10;
cout << *ptr1;
a)
5
b)
10
c)
The program will not compile
d)
Garbage value
|
|
Sonal Yadav answered |
The code snippet initializes an integer variable 'x' with the value 5. Two pointer variables 'ptr1' and 'ptr2' are declared and assigned the address of 'x' ('&x'). Then, 'ptr2' is assigned the value of 'ptr1'. Dereferencing 'ptr2' and assigning a value ('*ptr2 = 10';) modifies the value at the memory location pointed to by 'ptr2', which is 'x'. Therefore, the value of 'x' becomes 10. Printing '*ptr1' gives the output as 10.
What are smart pointers in C++?- a)Pointers that automatically deallocate memory when no longer needed
- b)Pointers that automatically resize themselves when needed
- c)Pointers that automatically update their values based on external changes
- d)Pointers that can point to multiple memory locations simultaneously
Correct answer is option 'D'. Can you explain this answer?
What are smart pointers in C++?
a)
Pointers that automatically deallocate memory when no longer needed
b)
Pointers that automatically resize themselves when needed
c)
Pointers that automatically update their values based on external changes
d)
Pointers that can point to multiple memory locations simultaneously
|
|
Sonal Yadav answered |
The code snippet attempts to assign the address of a function 'fun' to a function pointer variable 'ptr'. However, the syntax used for function pointers is incorrect. The correct syntax for declaring and initializing a function pointer is 'return_type (*ptr)(parameter_types)';. Therefore, the code will result in a compilation error.
What is a multidimensional dynamic array in C++?- a)An array with multiple dimensions that can change its size during runtime
- b)An array with multiple dimensions that are dynamically allocated
- c)An array with multiple dimensions that is allocated on the heap
- d)An array with multiple dimensions that is resized using the 'resize()' function
Correct answer is option 'B'. Can you explain this answer?
What is a multidimensional dynamic array in C++?
a)
An array with multiple dimensions that can change its size during runtime
b)
An array with multiple dimensions that are dynamically allocated
c)
An array with multiple dimensions that is allocated on the heap
d)
An array with multiple dimensions that is resized using the 'resize()' function
|
|
Sonal Yadav answered |
A multidimensional dynamic array in C++ is an array with multiple dimensions that are dynamically allocated. It allows for flexibility in the size of each dimension and can be resized using appropriate memory allocation and deallocation techniques.
When should function pointers be used in C++?- a)When passing a function as an argument to another function
- b)When returning a function from another function
- c)When implementing callbacks and event handlers
- d)All of the above
Correct answer is option 'B'. Can you explain this answer?
When should function pointers be used in C++?
a)
When passing a function as an argument to another function
b)
When returning a function from another function
c)
When implementing callbacks and event handlers
d)
All of the above
|
|
Sonal Yadav answered |
Function pointers are often used in scenarios where different functions need to be used based on runtime conditions or requirements. One common example is sorting an array of integers using a custom comparison function. By using a function pointer, the sorting algorithm can be made flexible to sort in ascending or descending order based on the provided comparison function.
What is the syntax to declare a function pointer in C++?- a)'void (*ptr)();'
- b)'voiid* ptr();'
- c)'voiid (*ptr)();
- d)'void* ptr();'
Correct answer is option 'D'. Can you explain this answer?
What is the syntax to declare a function pointer in C++?
a)
'void (*ptr)();'
b)
'voiid* ptr();'
c)
'voiid (*ptr)();
d)
'void* ptr();'
|
|
Sonal Yadav answered |
Pointers to member functions in C++ allow accessing and invoking member functions of objects through pointers. This provides a way to manipulate objects and their behaviors dynamically at runtime by using function pointers to member functions.
Consider the following code snippet:
#include <iostream>void processValues(int* ptr, int size) {
for (int i = 0; i < size; i++) {
std::cout << *ptr << " ";
ptr++;
}
}int main() {
int arr[] = {1, 2, 3, 4, 5};
processValues(arr + 1, 3); return 0;
}What will be the output of the above code?- a)2 3 4
- b)1 2 3
- c)3 4 5
- d)Error
Correct answer is option 'C'. Can you explain this answer?
Consider the following code snippet:
#include <iostream>
#include <iostream>
void processValues(int* ptr, int size) {
for (int i = 0; i < size; i++) {
std::cout << *ptr << " ";
ptr++;
}
}
for (int i = 0; i < size; i++) {
std::cout << *ptr << " ";
ptr++;
}
}
int main() {
int arr[] = {1, 2, 3, 4, 5};
processValues(arr + 1, 3);
int arr[] = {1, 2, 3, 4, 5};
processValues(arr + 1, 3);
return 0;
}
}
What will be the output of the above code?
a)
2 3 4
b)
1 2 3
c)
3 4 5
d)
Error
|
|
Sonal Yadav answered |
The 'processValues' function takes a pointer to the third element of the array ('arr + 1') and prints the next '3' elements. Therefore, the output will be 3 4 5.
What will be the output of the following code snippet?
#include <iostream>void changeValue(int* ptr) {
*ptr = 10;
}int main() {
int x = 5;
int* ptr = &x;
changeValue(ptr); std::cout << x; return 0;
}- a)5
- b)10
- c)Error
- d)Undefined behavior
Correct answer is option 'B'. Can you explain this answer?
What will be the output of the following code snippet?
#include <iostream>
#include <iostream>
void changeValue(int* ptr) {
*ptr = 10;
}
*ptr = 10;
}
int main() {
int x = 5;
int* ptr = &x;
changeValue(ptr);
int x = 5;
int* ptr = &x;
changeValue(ptr);
std::cout << x;
return 0;
}
}
a)
5
b)
10
c)
Error
d)
Undefined behavior
|
|
Sonal Yadav answered |
The 'changeValue' function modifies the value at the memory location pointed to by 'ptr'. In the 'main' function, 'x' is passed as an argument to 'changeValue', which changes the value of 'x' to 10. Therefore, the output will be 10.
What is the output of the following code snippet?
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr + 2;
cout << *ptr;- a)1
- b)2
- c)3
- d)4
Correct answer is option 'C'. Can you explain this answer?
What is the output of the following code snippet?
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr + 2;
cout << *ptr;
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr + 2;
cout << *ptr;
a)
1
b)
2
c)
3
d)
4
|
|
Sonal Yadav answered |
The code snippet initializes an integer array 'arr' with values {1, 2, 3, 4, 5}. A pointer variable 'ptr' is declared and assigned the address of the third element of the array ('arr + 2'). Dereferencing the pointer using the asterisk (*) operator ('*ptr') gives the value stored at the memory location pointed to by 'ptr', which is 3.
What is the size of a pointer in C++?- a)It depends on the type of variable it points to.
- b)4 bytes on 32-bit systems and 8 bytes on 64-bit systems.
- c)8 bytes on all systems.
- d)Pointers do not have a fixed size.
Correct answer is option 'A'. Can you explain this answer?
What is the size of a pointer in C++?
a)
It depends on the type of variable it points to.
b)
4 bytes on 32-bit systems and 8 bytes on 64-bit systems.
c)
8 bytes on all systems.
d)
Pointers do not have a fixed size.
|
|
Sonal Yadav answered |
The size of a pointer in C++ depends on the specific platform and the type of variable it points to. On most systems, a pointer typically has a fixed size of 4 bytes on 32-bit systems and 8 bytes on 64-bit systems.
What is a dynamic array in C++?- a)An array that can change its size during runtime
- b)An array with elements that can be dynamically allocated
- c)An array that is allocated on the heap
- d)An array that is resized using the 'resize()' function
Correct answer is option 'A'. Can you explain this answer?
What is a dynamic array in C++?
a)
An array that can change its size during runtime
b)
An array with elements that can be dynamically allocated
c)
An array that is allocated on the heap
d)
An array that is resized using the 'resize()' function
|
|
Sonal Yadav answered |
A dynamic array in C++ is an array that can change its size during runtime. It is allocated using the 'new' operator and deallocated using the 'delete[]' operator. This allows the array to have a flexible size based on program requirements.
What does the "void" keyword represent in C++?- a)It indicates that the function returns nothing.
- b)It is used to declare a variable of type "void."
- c)It is used to define a generic pointer.
- d)It represents an error in the code.
Correct answer is option 'A'. Can you explain this answer?
What does the "void" keyword represent in C++?
a)
It indicates that the function returns nothing.
b)
It is used to declare a variable of type "void."
c)
It is used to define a generic pointer.
d)
It represents an error in the code.
|
|
Sonal Yadav answered |
In C++, the "void" keyword is used to indicate that a function does not return a value. It is typically used when the function is meant to perform an action or when the return value is not required.
What is the size of a pointer in C++?- a)It depends on the type of the variable the pointer points to
- b)It is always 4 bytes
- c)It is always 8 bytes
- d)It is always 2 bytes
Correct answer is option 'A'. Can you explain this answer?
What is the size of a pointer in C++?
a)
It depends on the type of the variable the pointer points to
b)
It is always 4 bytes
c)
It is always 8 bytes
d)
It is always 2 bytes
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Sonal Yadav answered |
The size of a pointer in C++ depends on the underlying data type it points to. For example, the size of an 'int*' pointer is typically 4 bytes on a 32-bit system and 8 bytes on a 64-bit system.
What is the output of the following code snippet?
#include <memory>int main() {
std::unique_ptr<int> ptr(new int(5));
std::cout << *ptr;
return 0;
}- a)0
- b)5
- c)The program will not compile
- d)Garbage value
Correct answer is option 'C'. Can you explain this answer?
What is the output of the following code snippet?
#include <memory>
#include <memory>
int main() {
std::unique_ptr<int> ptr(new int(5));
std::cout << *ptr;
return 0;
}
std::unique_ptr<int> ptr(new int(5));
std::cout << *ptr;
return 0;
}
a)
0
b)
5
c)
The program will not compile
d)
Garbage value
|
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Sonal Yadav answered |
Function pointers in C++ provide a way to pass functions as arguments to other functions. This allows for increased flexibility and modularity in program design by enabling the use of different functions based on runtime conditions or requirements.
What is the output of the following code snippet?
#include <iostream>void printValues(int* ptr) {
for (int i = 0; i < 5; i++) {
std::cout << *ptr << " ";
ptr++;
}
}int main() {
int arr[] = {1, 2, 3, 4, 5};
printValues(arr); return 0;
}- a)1 2 3 4 5
- b)5 4 3 2 1
- c)1 1 1 1 1
- d)Error
Correct answer is option 'A'. Can you explain this answer?
What is the output of the following code snippet?
#include <iostream>
#include <iostream>
void printValues(int* ptr) {
for (int i = 0; i < 5; i++) {
std::cout << *ptr << " ";
ptr++;
}
}
for (int i = 0; i < 5; i++) {
std::cout << *ptr << " ";
ptr++;
}
}
int main() {
int arr[] = {1, 2, 3, 4, 5};
printValues(arr);
int arr[] = {1, 2, 3, 4, 5};
printValues(arr);
return 0;
}
}
a)
1 2 3 4 5
b)
5 4 3 2 1
c)
1 1 1 1 1
d)
Error
|
|
Sonal Yadav answered |
The 'printValues' function takes a pointer to the first element of the array and prints the values by incrementing the pointer. In the 'main' function, 'printValues' is called with 'arr' as an argument, which is the pointer to the first element of the array. Therefore, the output will be 1 2 3 4 5.
Which keyword is used to allocate memory dynamically in C++?- a)new
- b)malloc
- c)alloc
- d)dynamic
Correct answer is option 'A'. Can you explain this answer?
Which keyword is used to allocate memory dynamically in C++?
a)
new
b)
malloc
c)
alloc
d)
dynamic
|
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Sonal Yadav answered |
The 'new' keyword is used to allocate memory dynamically in C++. It is used to allocate memory for objects and data structures during runtime.
Consider the following code snippet:
#include <iostream>void printArray(int arr[]) {
std::cout << arr[0];
}int main() {
int arr[5] = {1, 2, 3, 4, 5};
printArray(arr); return 0;
}
What will be the output of the above code?- a)1
- b)2
- c)3
- d)Error
Correct answer is option 'D'. Can you explain this answer?
Consider the following code snippet:
#include <iostream>
#include <iostream>
void printArray(int arr[]) {
std::cout << arr[0];
}
std::cout << arr[0];
}
int main() {
int arr[5] = {1, 2, 3, 4, 5};
printArray(arr);
int arr[5] = {1, 2, 3, 4, 5};
printArray(arr);
return 0;
}
What will be the output of the above code?
}
What will be the output of the above code?
a)
1
b)
2
c)
3
d)
Error
|
|
Sonal Yadav answered |
The function 'printArray' expects an array as an argument, but the size of the array is not specified. This code will lead to an error because the size of the array is required for proper access.
What is the output of the following code snippet?
int* ptr = new int[5];
ptr[2] = 10;
cout << ptr[2];
delete[] ptr;- a)0
- b)10
- c)The program will not compile
- d)Garbage value
Correct answer is option 'B'. Can you explain this answer?
What is the output of the following code snippet?
int* ptr = new int[5];
ptr[2] = 10;
cout << ptr[2];
delete[] ptr;
int* ptr = new int[5];
ptr[2] = 10;
cout << ptr[2];
delete[] ptr;
a)
0
b)
10
c)
The program will not compile
d)
Garbage value
|
|
Sonal Yadav answered |
The code snippet dynamically allocates an integer array 'ptr' of size 5 using the new operator. The 'ptr[2] = 10'; statement assigns the value 10 to the element at index 2 of the array. Finally, 'cout << ptr[2]'; prints the value at index 2 of the array, which is 10. It is important to note that the dynamically allocated array is deallocated using the 'delete[]' operator to avoid memory leaks.
What is the output of the following code snippet?
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr;
cout << ptr[2];- a)1
- b)2
- c)3
- d)4
Correct answer is option 'C'. Can you explain this answer?
What is the output of the following code snippet?
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr;
cout << ptr[2];
int arr[] = {1, 2, 3, 4, 5};
int* ptr = arr;
cout << ptr[2];
a)
1
b)
2
c)
3
d)
4
|
|
Sonal Yadav answered |
The code snippet initializes an integer array 'arr' with values {1, 2, 3, 4, 5}. A pointer variable 'ptr' is declared and assigned the address of the first element of the array ('arr'). The expression 'ptr[2]' is equivalent to '*(ptr + 2)', which accesses the value at the memory location two positions ahead of 'ptr', resulting in the value 3.
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C++ for EmSAT Achieve
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