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All questions of Dynamic Allocation for Software Development Exam
What will be the output of the following code snippet?
```cpp
int** matrix = new int*[3];
for (int i = 0; i < 3; i++) {
matrix[i] = new int[2];
for (int j = 0; j < 2; j++) {
matrix[i][j] = i * j;
}
}
cout << matrix[2][1] << endl;
```- a)0
- b)1
- c)2
- d)Compile error
Correct answer is option 'C'. Can you explain this answer?
What will be the output of the following code snippet?
```cpp
int** matrix = new int*[3];
for (int i = 0; i < 3; i++) {
matrix[i] = new int[2];
for (int j = 0; j < 2; j++) {
matrix[i][j] = i * j;
}
}
cout << matrix[2][1] << endl;
```
```cpp
int** matrix = new int*[3];
for (int i = 0; i < 3; i++) {
matrix[i] = new int[2];
for (int j = 0; j < 2; j++) {
matrix[i][j] = i * j;
}
}
cout << matrix[2][1] << endl;
```
a)
0
b)
1
c)
2
d)
Compile error
|
Code Nation answered |
The matrix is a dynamic 2D array with 3 rows and 2 columns. The value at matrix[2][1] is 2 (2 * 1).
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What is the output of the following code?
#include <iostream>
using namespace std;int main() {
int* ptr = new int(5);
cout << *ptr << endl;
delete ptr;
return 0;
}- a)5
- b)0
- c)Garbage value
- d)Compilation error
Correct answer is option 'A'. Can you explain this answer?
What is the output of the following code?
#include <iostream>
using namespace std;
#include <iostream>
using namespace std;
int main() {
int* ptr = new int(5);
cout << *ptr << endl;
delete ptr;
return 0;
}
int* ptr = new int(5);
cout << *ptr << endl;
delete ptr;
return 0;
}
a)
5
b)
0
c)
Garbage value
d)
Compilation error
|
Thangaperumal. answered |
Tamilnadu government THANGAPERUMAL BURPT3397R
Which operator is used to deallocate memory allocated for a single object in C++?- a)delete
- b)dealloc
- c)free
- d)remove
Correct answer is option 'A'. Can you explain this answer?
Which operator is used to deallocate memory allocated for a single object in C++?
a)
delete
b)
dealloc
c)
free
d)
remove
|
Nilotpal Jain answered |
delete operator in C++
The delete operator in C++ is used to deallocate memory that was previously allocated for a single object using the new operator. It is important to free up memory that is no longer needed in order to prevent memory leaks and improve the efficiency of your program.
How to use the delete operator:
- To deallocate memory for a single object, you simply use the delete operator followed by the pointer to the object that was allocated using new.
- For example, if you allocated memory for an object of class MyClass using new, you can deallocate the memory using delete as follows: delete myClassObject;
Importance of using the delete operator:
- Failure to deallocate memory can lead to memory leaks, where memory is allocated but never released, causing your program to consume more and more memory over time.
- By using the delete operator properly, you can ensure that memory is released when it is no longer needed, improving the overall performance and stability of your program.
In conclusion, the delete operator is essential in C++ for deallocating memory allocated for a single object. It is important to use it correctly to prevent memory leaks and optimize the memory usage of your program.
The delete operator in C++ is used to deallocate memory that was previously allocated for a single object using the new operator. It is important to free up memory that is no longer needed in order to prevent memory leaks and improve the efficiency of your program.
How to use the delete operator:
- To deallocate memory for a single object, you simply use the delete operator followed by the pointer to the object that was allocated using new.
- For example, if you allocated memory for an object of class MyClass using new, you can deallocate the memory using delete as follows: delete myClassObject;
Importance of using the delete operator:
- Failure to deallocate memory can lead to memory leaks, where memory is allocated but never released, causing your program to consume more and more memory over time.
- By using the delete operator properly, you can ensure that memory is released when it is no longer needed, improving the overall performance and stability of your program.
In conclusion, the delete operator is essential in C++ for deallocating memory allocated for a single object. It is important to use it correctly to prevent memory leaks and optimize the memory usage of your program.
Which keyword is used to allocate dynamic memory in C++?- a)malloc
- b)calloc
- c)new
- d)allocate
Correct answer is option 'C'. Can you explain this answer?
Which keyword is used to allocate dynamic memory in C++?
a)
malloc
b)
calloc
c)
new
d)
allocate
|
CodeNation answered |
The new keyword is used to dynamically allocate memory in C++.
Which operator is used to dynamically allocate memory for a single object in C++?- a)new
- b)malloc
- c)alloc
- d)allocate
Correct answer is option 'A'. Can you explain this answer?
Which operator is used to dynamically allocate memory for a single object in C++?
a)
new
b)
malloc
c)
alloc
d)
allocate
|
|
CodeNation answered |
The new operator is used to dynamically allocate memory for a single object in C++.
What is the output of the following code?
```cpp
#include <iostream>
using namespace std;int getValue() {
static int num = 0;
num++;
return num;
}int main() {
cout << getValue() << endl;
cout << getValue() << endl;
cout << getValue() << endl;
return 0;
}
```- a)1, 2, 3
- b)1, 1, 1
- c)1, 1, 2
- d)Compilation error
Correct answer is option 'C'. Can you explain this answer?
What is the output of the following code?
```cpp
#include <iostream>
using namespace std;
```cpp
#include <iostream>
using namespace std;
int getValue() {
static int num = 0;
num++;
return num;
}
static int num = 0;
num++;
return num;
}
int main() {
cout << getValue() << endl;
cout << getValue() << endl;
cout << getValue() << endl;
return 0;
}
```
cout << getValue() << endl;
cout << getValue() << endl;
cout << getValue() << endl;
return 0;
}
```
a)
1, 2, 3
b)
1, 1, 1
c)
1, 1, 2
d)
Compilation error
|
Qudrat Chauhan answered |
The function 'getValue' has a static local variable 'num' that is initialized to 0. Each time the function is called, 'num' is incremented by 1 and returned. In the main function, 'getValue' is called three times, resulting in the sequence 1, 2, 3 being printed.
What happens if we forget to deallocate the memory allocated using new?- a)The program crashes
- b)The memory is automatically deallocated
- c)Memory leak occurs
- d)The program shows a memory error
Correct answer is option 'C'. Can you explain this answer?
What happens if we forget to deallocate the memory allocated using new?
a)
The program crashes
b)
The memory is automatically deallocated
c)
Memory leak occurs
d)
The program shows a memory error
|
|
CodeNation answered |
If we forget to deallocate the memory allocated using new, a memory leak
Which operator is used to dynamically allocate memory in C++?- a)new
- b)malloc
- c)alloc
- d)allocate
Correct answer is option 'A'. Can you explain this answer?
Which operator is used to dynamically allocate memory in C++?
a)
new
b)
malloc
c)
alloc
d)
allocate
|
Hackers World answered |
The 'new' operator is used for dynamic memory allocation in C++. It returns the address of the memory block allocated.
What will be the output of the following code snippet?
void changeArray(int arr[]) {
arr[0] = 5;
}int main() {
int arr[3] = {1, 2, 3};
changeArray(arr);
cout << arr[0] << endl;
return 0;
}- a)1
- b)5
- c)Compile error
- d)Runtime error
Correct answer is option 'A'. Can you explain this answer?
What will be the output of the following code snippet?
void changeArray(int arr[]) {
arr[0] = 5;
}
void changeArray(int arr[]) {
arr[0] = 5;
}
int main() {
int arr[3] = {1, 2, 3};
changeArray(arr);
cout << arr[0] << endl;
return 0;
}
int arr[3] = {1, 2, 3};
changeArray(arr);
cout << arr[0] << endl;
return 0;
}
a)
1
b)
5
c)
Compile error
d)
Runtime error
|
Ritesh Kumar Shivhare answered |
When the array 'arr' is passed to the function 'changeArray', it is passed as a pointer. Thus, modifying the array inside the function will also affect the original array.
What will be the output of the following code snippet?
int* ptr = new int[3]{1, 2, 3};
cout << *ptr << endl;
delete[] ptr;
cout << *ptr << endl;- a)1, 1
- b)1, garbage value
- c)1, runtime error
- d)Compile error
Correct answer is option 'C'. Can you explain this answer?
What will be the output of the following code snippet?
int* ptr = new int[3]{1, 2, 3};
cout << *ptr << endl;
delete[] ptr;
cout << *ptr << endl;
int* ptr = new int[3]{1, 2, 3};
cout << *ptr << endl;
delete[] ptr;
cout << *ptr << endl;
a)
1, 1
b)
1, garbage value
c)
1, runtime error
d)
Compile error
|
|
CodeNation answered |
The first cout statement will print 1, but after deleting the memory using delete[], accessing the memory using the pointer will result in a runtime error.
What will be the output of the following code snippet?
void changeValue(int* ptr) {
*ptr = 20;
}int main() {
int value = 10;
changeValue(&value);
cout << value << endl;
return 0;
}- a)10
- b)20
- c)Compile error
- d)Runtime error
Correct answer is option 'B'. Can you explain this answer?
What will be the output of the following code snippet?
void changeValue(int* ptr) {
*ptr = 20;
}
void changeValue(int* ptr) {
*ptr = 20;
}
int main() {
int value = 10;
changeValue(&value);
cout << value << endl;
return 0;
}
int value = 10;
changeValue(&value);
cout << value << endl;
return 0;
}
a)
10
b)
20
c)
Compile error
d)
Runtime error
|
|
Ritesh Kumar Shivhare answered |
The function 'changeValue' accepts a pointer and modifies the value at the memory address pointed to by the pointer. So, the value of 'value' changes to 20.
What will be the output of the following code snippet?
```cpp
void changeValue(int** matrix) {
matrix[0] = new int[2]{5, 5};
}int main() {
int** matrix = new int*[2];
for (int i = 0; i < 2; i++) {
matrix[i] = new int[2]{1, 1};
}
changeValue(matrix);
cout << matrix[0][1] << endl;
return 0;
}
```- a)0
- b)1
- c)5
- d)Compile error
Correct answer is option 'C'. Can you explain this answer?
What will be the output of the following code snippet?
```cpp
void changeValue(int** matrix) {
matrix[0] = new int[2]{5, 5};
}
void changeValue(int** matrix) {
matrix[0] = new int[2]{5, 5};
}
int main() {
int** matrix = new int*[2];
for (int i = 0; i < 2; i++) {
matrix[i] = new int[2]{1, 1};
}
changeValue(matrix);
cout << matrix[0][1] << endl;
return 0;
}
```
int** matrix = new int*[2];
for (int i = 0; i < 2; i++) {
matrix[i] = new int[2]{1, 1};
}
changeValue(matrix);
cout << matrix[0][1] << endl;
return 0;
}
```
a)
0
b)
1
c)
5
d)
Compile error
|
Tanuja Mishra answered |
The function 'changeValue' receives a pointer to a 2D array. It dynamically allocates a new row and assigns the values 5 and 5. Thus, matrix[0][1] becomes 5.
What is the output of the following code?
#include <iostream>
using namespace std;#define SQUARE(x) x*xint main() {
int num = 5;
cout << SQUARE(num + 1) << endl;
return 0;
}- a)10
- b)30
- c)36
- d)Compilation error
Correct answer is option 'C'. Can you explain this answer?
What is the output of the following code?
#include <iostream>
using namespace std;
#include <iostream>
using namespace std;
#define SQUARE(x) x*x
int main() {
int num = 5;
cout << SQUARE(num + 1) << endl;
return 0;
}
int num = 5;
cout << SQUARE(num + 1) << endl;
return 0;
}
a)
10
b)
30
c)
36
d)
Compilation error
|
Neeraj Singh Kushwah answered |
The code defines a macro SQUARE that squares its argument. In the main function, the expression SQUARE(num + 1) is expanded as (num + 1 * num + 1) due to the absence of parentheses in the macro definition. This leads to the expression being evaluated as (5 + 1 * 5 + 1) = 11. The output is 11 * 11 = 121, which is printed.
Which keyword is used to declare a constant variable in C++?- a)final
- b)const
- c)constant
- d)var
Correct answer is option 'B'. Can you explain this answer?
Which keyword is used to declare a constant variable in C++?
a)
final
b)
const
c)
constant
d)
var
|
|
Neeraj Singh Kushwah answered |
The 'const' keyword is used to declare a constant variable in C++. The value of a constant variable cannot be modified once it is assigned.
What is the output of the following code?
#include <iostream>
using namespace std;int sum(int x = 2, int y = 3);int main() {
cout << sum(5) << endl;
return 0;
}int sum(int x, int y) {
return x + y;
}- a)2
- b)3
- c)5
- d)8
Correct answer is option 'C'. Can you explain this answer?
What is the output of the following code?
#include <iostream>
using namespace std;
#include <iostream>
using namespace std;
int sum(int x = 2, int y = 3);
int main() {
cout << sum(5) << endl;
return 0;
}
cout << sum(5) << endl;
return 0;
}
int sum(int x, int y) {
return x + y;
}
return x + y;
}
a)
2
b)
3
c)
5
d)
8
|
Startup Central Classes answered |
The 'sum' function is declared with default parameters. In the main function, only the first argument is provided, and the default value for the second argument is used. Thus, the function call sum(5) is equivalent to sum(5, 3), resulting in 5 + 3 = 8. The output is 8.
What is the output of the following code?
#include <iostream>
using namespace std;static int count = 0;int main() {
cout << count++ << endl;
count++;
cout << count << endl;
return 0;
}- a)0, 2
- b)0, 1
- c)1, 1
- d)0, 0
Correct answer is option 'A'. Can you explain this answer?
What is the output of the following code?
#include <iostream>
using namespace std;
#include <iostream>
using namespace std;
static int count = 0;
int main() {
cout << count++ << endl;
count++;
cout << count << endl;
return 0;
}
cout << count++ << endl;
count++;
cout << count << endl;
return 0;
}
a)
0, 2
b)
0, 1
c)
1, 1
d)
0, 0
|
|
Startup Central Classes answered |
The 'count' variable is declared as static, which means it retains its value between function calls. In the main function, count is printed with the post-increment operator, resulting in 0 being printed. Then, count is incremented twice, resulting in its value being 2 when printed again.
What will be the output of the following code snippet?
int a = 10;
int& ref = a;
int* ptr = &ref;
cout << *ptr << endl;- a)10
- b)Compile error
- c)Runtime error
- d)Garbage value
Correct answer is option 'A'. Can you explain this answer?
What will be the output of the following code snippet?
int a = 10;
int& ref = a;
int* ptr = &ref;
cout << *ptr << endl;
int a = 10;
int& ref = a;
int* ptr = &ref;
cout << *ptr << endl;
a)
10
b)
Compile error
c)
Runtime error
d)
Garbage value
|
|
Ritesh Kumar Shivhare answered |
The pointer 'ptr' is pointing to the address of 'a', so dereferencing the pointer will give the value 10.
What will be the output of the following code snippet?
```cpp
int** matrix = new int*[2];
for (int i = 0; i < 2; i++) {
matrix[i] = new int[3];
for (int j = 0; j < 3; j++) {
matrix[i][j] = i + j;
}
}
cout << matrix[1][2] << endl;
```- a)0
- b)2
- c)3
- d)Compile error
Correct answer is option 'B'. Can you explain this answer?
What will be the output of the following code snippet?
```cpp
int** matrix = new int*[2];
for (int i = 0; i < 2; i++) {
matrix[i] = new int[3];
for (int j = 0; j < 3; j++) {
matrix[i][j] = i + j;
}
}
cout << matrix[1][2] << endl;
```
```cpp
int** matrix = new int*[2];
for (int i = 0; i < 2; i++) {
matrix[i] = new int[3];
for (int j = 0; j < 3; j++) {
matrix[i][j] = i + j;
}
}
cout << matrix[1][2] << endl;
```
a)
0
b)
2
c)
3
d)
Compile error
|
|
Code Nation answered |
The matrix is a dynamic 2D array with 2 rows and 3 columns. The value at matrix[1][2] is 2 (1 + 2).
What is the purpose of the delete keyword in C++?- a)To deallocate memory allocated using new
- b)To free memory allocated using malloc
- c)To remove an element from an array
- d)To release resources used by a function
Correct answer is option 'A'. Can you explain this answer?
What is the purpose of the delete keyword in C++?
a)
To deallocate memory allocated using new
b)
To free memory allocated using malloc
c)
To remove an element from an array
d)
To release resources used by a function
|
|
CodeNation answered |
The delete keyword is used to deallocate memory that was allocated using new.
What will be the output of the following code snippet?
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)Compile error
Correct answer is option 'C'. Can you explain this answer?
What will be the output of the following code snippet?
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)
Compile error
|
|
CodeNation answered |
The first cout statement will print 5, but after deleting the memory using delete, accessing the memory using the pointer will result in a runtime error.
What is the output of the following code?
```cpp
#include <iostream>
using namespace std;inline int multiply(int a, int b) {
return a * b;
}int main() {
int result = multiply(2, 3) + multiply(4, 5);
cout << result << endl;
return 0;
}
```- a)14
- b)26
- c)22
- d)Compilation error
Correct answer is option 'C'. Can you explain this answer?
What is the output of the following code?
```cpp
#include <iostream>
using namespace std;
```cpp
#include <iostream>
using namespace std;
inline int multiply(int a, int b) {
return a * b;
}
return a * b;
}
int main() {
int result = multiply(2, 3) + multiply(4, 5);
cout << result << endl;
return 0;
}
```
int result = multiply(2, 3) + multiply(4, 5);
cout << result << endl;
return 0;
}
```
a)
14
b)
26
c)
22
d)
Compilation error
|
Yogesh Dwivedi answered |
The 'inline' keyword suggests the compiler to replace the function call to 'multiply' with its body. Thus, the expression becomes 2 * 3 + 4 * 5, which simplifies to 6 + 20 = 26.
What is the output of the following code?
#include <iostream>
using namespace std;int main() {
const int x = 5;
int* ptr = (int*)&x;
*ptr = 10;
cout << x << endl;
return 0;
}- a)5
- b)10
- c)Runtime error
- d)Compilation error
Correct answer is option 'A'. Can you explain this answer?
What is the output of the following code?
#include <iostream>
using namespace std;
#include <iostream>
using namespace std;
int main() {
const int x = 5;
int* ptr = (int*)&x;
*ptr = 10;
cout << x << endl;
return 0;
}
const int x = 5;
int* ptr = (int*)&x;
*ptr = 10;
cout << x << endl;
return 0;
}
a)
5
b)
10
c)
Runtime error
d)
Compilation error
|
|
Startup Central Classes answered |
The 'const' keyword is used to declare a constant variable 'x'. However, the code tries to modify the value of 'x' by using a pointer and assigning a new value through it. Modifying a constant variable results in undefined behavior, but in this case, the value of 'x' remains unchanged as it is optimized by the compiler.
Which keyword is used to deallocate memory in C++?- a)free
- b)deallocate
- c)delete
- d)remove
Correct answer is option 'C'. Can you explain this answer?
Which keyword is used to deallocate memory in C++?
a)
free
b)
deallocate
c)
delete
d)
remove
|
|
Neeraj Singh Kushwah answered |
The 'delete' keyword is used to deallocate dynamically allocated memory in C++. It frees the memory block and makes it available for future allocations.
What is the output of the following code?
```cpp
#include <iostream>
using namespace std;#define MAX(a, b) ((a) > (b) ? (a) : (b))int main() {
int num1 = 10, num2 = 5, num3 = 8;
int maxNum = MAX(num1++, num2++);
cout << maxNum << endl;
return 0;
}
```- a)10
- b)5
- c)8
- d)Compilation error
Correct answer is option 'B'. Can you explain this answer?
What is the output of the following code?
```cpp
#include <iostream>
using namespace std;
```cpp
#include <iostream>
using namespace std;
#define MAX(a, b) ((a) > (b) ? (a) : (b))
int main() {
int num1 = 10, num2 = 5, num3 = 8;
int maxNum = MAX(num1++, num2++);
cout << maxNum << endl;
return 0;
}
```
int num1 = 10, num2 = 5, num3 = 8;
int maxNum = MAX(num1++, num2++);
cout << maxNum << endl;
return 0;
}
```
a)
10
b)
5
c)
8
d)
Compilation error
|
|
Yogesh Dwivedi answered |
The macro MAX is defined to return the maximum value between two arguments.
In the main function, num1++ and num2++ are passed as arguments.
Since the macro directly substitutes the arguments, the expressions become num1++ > num2++ ? num1++ : num2++, resulting in 10 > 5 ? 10++ : 5++, which evaluates to 5.
In the main function, num1++ and num2++ are passed as arguments.
Since the macro directly substitutes the arguments, the expressions become num1++ > num2++ ? num1++ : num2++, resulting in 10 > 5 ? 10++ : 5++, which evaluates to 5.
What will be the output of the following code snippet?```cpp
void changeValue(int*** matrix) {
matrix[1][0][0] = 10;
}int main() {
int*** matrix = new int**[2];
for (int i = 0; i < 2; i++) {
matrix[i] = new int*[2];
for (int j = 0; j < 2; j++) {
matrix[i][j] = new int[2]{i, j};
}
}
changeValue(matrix);
cout << matrix[1][0][0] << endl;
return 0;
}
```- a)0
- b)1
- c)10
- d)Compile error
Correct answer is option 'C'. Can you explain this answer?
What will be the output of the following code snippet?
```cpp
void changeValue(int*** matrix) {
matrix[1][0][0] = 10;
}
void changeValue(int*** matrix) {
matrix[1][0][0] = 10;
}
int main() {
int*** matrix = new int**[2];
for (int i = 0; i < 2; i++) {
matrix[i] = new int*[2];
for (int j = 0; j < 2; j++) {
matrix[i][j] = new int[2]{i, j};
}
}
changeValue(matrix);
cout << matrix[1][0][0] << endl;
return 0;
}
```
int*** matrix = new int**[2];
for (int i = 0; i < 2; i++) {
matrix[i] = new int*[2];
for (int j = 0; j < 2; j++) {
matrix[i][j] = new int[2]{i, j};
}
}
changeValue(matrix);
cout << matrix[1][0][0] << endl;
return 0;
}
```
a)
0
b)
1
c)
10
d)
Compile error
|
Codebreakers answered |
The function 'changeValue' receives a pointer to a 3D array. Modifying the value at matrix[1][0][0] inside the function changes the original array as well.
What is the output of the following code?
```cpp
#include <iostream>
using namespace std;int main() {
const int arr[] = {1, 2, 3};
int* ptr = const_cast<int*>(arr);
*ptr = 10;
cout << arr[0] << endl;
return 0;
}
```- a)10
- b)1
- c)Runtime error
- d)Compilation error
Correct answer is option 'C'. Can you explain this answer?
What is the output of the following code?
```cpp
#include <iostream>
using namespace std;
```cpp
#include <iostream>
using namespace std;
int main() {
const int arr[] = {1, 2, 3};
int* ptr = const_cast<int*>(arr);
*ptr = 10;
cout << arr[0] << endl;
return 0;
}
```
const int arr[] = {1, 2, 3};
int* ptr = const_cast<int*>(arr);
*ptr = 10;
cout << arr[0] << endl;
return 0;
}
```
a)
10
b)
1
c)
Runtime error
d)
Compilation error
|
Vikram Singh Baghel answered |
The code tries to modify a constant array 'arr' by using const_cast to remove the const qualifier. Modifying a constant object results in undefined behavior, which in this case leads to a runtime error.
What will be the output of the following code snippet?```cpp
void changeValue(int** matrix) {
matrix[0][1] = 5;
}int main() {
int** matrix = new int*[2];
for (int i = 0; i < 2; i++) {
matrix[i] = new int[2];
for (int j = 0; j < 2; j++) {
matrix[i][j] = i + j;
}
}
changeValue(matrix);
cout << matrix[0][1] << endl;
return 0;
}
```- a)0
- b)2
- c)5
- d)Compile error
Correct answer is option 'C'. Can you explain this answer?
What will be the output of the following code snippet?
```cpp
void changeValue(int** matrix) {
matrix[0][1] = 5;
}
void changeValue(int** matrix) {
matrix[0][1] = 5;
}
int main() {
int** matrix = new int*[2];
for (int i = 0; i < 2; i++) {
matrix[i] = new int[2];
for (int j = 0; j < 2; j++) {
matrix[i][j] = i + j;
}
}
changeValue(matrix);
cout << matrix[0][1] << endl;
return 0;
}
```
int** matrix = new int*[2];
for (int i = 0; i < 2; i++) {
matrix[i] = new int[2];
for (int j = 0; j < 2; j++) {
matrix[i][j] = i + j;
}
}
changeValue(matrix);
cout << matrix[0][1] << endl;
return 0;
}
```
a)
0
b)
2
c)
5
d)
Compile error
|
|
Tanuja Mishra answered |
The function 'changeValue' receives a pointer to a 2D array. Modifying the value at matrix[0][1] inside the function changes the original array as well.
What is the purpose of using macros in C++?- a)To define constants
- b)To declare variables
- c)To implement loops
- d)To allocate memory
Correct answer is option 'A'. Can you explain this answer?
What is the purpose of using macros in C++?
a)
To define constants
b)
To declare variables
c)
To implement loops
d)
To allocate memory
|
|
Neeraj Singh Kushwah answered |
Macros in C++ are used to define constants and perform text substitution during compilation.
What does the keyword "inline" indicate in a function declaration?- a)The function can be called from any file
- b)The function is defined in a separate file
- c)The function will be replaced with its body during compilation
- d)The function is executed in parallel
Correct answer is option 'C'. Can you explain this answer?
What does the keyword "inline" indicate in a function declaration?
a)
The function can be called from any file
b)
The function is defined in a separate file
c)
The function will be replaced with its body during compilation
d)
The function is executed in parallel
|
|
Neeraj Singh Kushwah answered |
The 'inline' keyword in a function declaration suggests the compiler to replace the function call with its body at the call site, reducing function call overhead.
What is the output of the following code?
```cpp
#include <iostream>
using namespace std;const int num = 5;void increment() {
const int num = 10;
cout << ++num << endl;
}int main() {
increment();
cout << num << endl;
return 0;
}
```- a)11, 5
- b)6, 5
- c)11, 10
- d)6, 10
Correct answer is option 'A'. Can you explain this answer?
What is the output of the following code?
```cpp
#include <iostream>
using namespace std;
```cpp
#include <iostream>
using namespace std;
const int num = 5;
void increment() {
const int num = 10;
cout << ++num << endl;
}
const int num = 10;
cout << ++num << endl;
}
int main() {
increment();
cout << num << endl;
return 0;
}
```
increment();
cout << num << endl;
return 0;
}
```
a)
11, 5
b)
6, 5
c)
11, 10
d)
6, 10
|
|
Qudrat Chauhan answered |
The function increment() has a local constant variable 'num' with a value of 10. When incremented (++num), its value becomes 11, which is printed. In the main function, the global constant variable 'num' with a value of 5 is printed. Constants with the same name in different scopes are treated as separate entities.
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