What happens if we run the following code?12345678#include <iostream>int main() { int* ptr = new int; ptr = new int; delete ptr; return 0;}Marks : 1Negative Marks : 0Answer hereThe code will not compile due to a syntax error.The code will compile and execute without any issues.The code will compile but result in a memory leak.The code will compile but result in an allocation failure.
Question
What happens if we run the following code?12345678#include <iostream>int main() { int* ptr = new int; ptr = new int; delete ptr; return 0;}Marks : 1Negative Marks : 0Answer hereThe code will not compile due to a syntax error.The code will compile and execute without any issues.The code will compile but result in a memory leak.The code will compile but result in an allocation failure.
Solution
The code will compile but result in a memory leak.
Here's why:
- The pointer 'ptr' is allocated memory using 'new int'.
- Then, without freeing the previously allocated memory, 'ptr' is again allocated memory using 'new int'.
- After that, 'delete ptr' is used, which only frees the memory allocated in the second step.
- The memory allocated in the first step is not freed, resulting in a memory leak.
Similar Questions
What happens if we compile the following code?
What will be the output of the following program?1234567891011121314151617181920#include <stdio.h>void one();void two();void one() { for(int i = 0; i < 2; i++){ two(); }}void two() { printf("TWO ");}int main(){ one(); two(); return 0;}
What happens if we compile the following code?12345678#include <stdio.h>int main() { int num = 42; int *ptr = # printf("%d", *ptr); return 0;}
What will print out when the following code runs?
What happens if we run the following code?12345678#include <iostream>int main() { int* ptr = new int; ptr = new int; delete ptr; return 0;}Marks : 1Negative Marks : 0Answer hereThe code will not compile due to a syntax error.The code will compile and execute without any issues.The code will compile but result in a memory leak.The code will compile but result in an allocation failure.
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