EmSAT Achieve Exam > EmSAT Achieve Questions > Consider the following code snippet:#include ...
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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
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Consider the following code snippet:#include <iostream>void chan...
Explanation:
- changeValue Function:
- The function `changeValue` takes a pointer to an integer as a parameter.
- It dereferences the pointer and assigns the value 20 to the integer pointed to by `ptr`.
- It then assigns `nullptr` to the pointer `ptr`, making it point to nothing.
- Main Function:
- In the `main` function, an integer `x` is declared and initialized to 10.
- A pointer `ptr` is created and assigned the address of `x`.
- The `changeValue` function is called with `ptr` as an argument.
- Output:
- After the `changeValue` function is called, the value of `x` is changed to 20.
- However, the pointer `ptr` inside the `main` function still points to the address of `x` and is not affected by the assignment of `nullptr` inside the `changeValue` function.
- Therefore, the expression `(ptr == nullptr)` in the `main` function will evaluate to false, and the output will be 0.
Therefore, the output of the above code will be:
a) 0
- changeValue Function:
- The function `changeValue` takes a pointer to an integer as a parameter.
- It dereferences the pointer and assigns the value 20 to the integer pointed to by `ptr`.
- It then assigns `nullptr` to the pointer `ptr`, making it point to nothing.
- Main Function:
- In the `main` function, an integer `x` is declared and initialized to 10.
- A pointer `ptr` is created and assigned the address of `x`.
- The `changeValue` function is called with `ptr` as an argument.
- Output:
- After the `changeValue` function is called, the value of `x` is changed to 20.
- However, the pointer `ptr` inside the `main` function still points to the address of `x` and is not affected by the assignment of `nullptr` inside the `changeValue` function.
- Therefore, the expression `(ptr == nullptr)` in the `main` function will evaluate to false, and the output will be 0.
Therefore, the output of the above code will be:
a) 0
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Community Answer
Consider the following code snippet:#include <iostream>void chan...
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'.
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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)0b)1c)Errord)Undefined behaviorCorrect answer is option 'A'. Can you explain this answer?
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