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C | Dynamic Memory Allocation | Question 3
  • Difficulty Level : Easy
  • Last Updated : 09 Feb, 2013

Output?




# include<stdio.h>
# include<stdlib.h>
   
void fun(int *a)
{
    a = (int*)malloc(sizeof(int));
}
   
int main()
{
    int *p;
    fun(p);
    *p = 6;
    printf("%d\n",*p);
    return(0);
}

(A) May not work
(B) Works and prints 6


Answer: (A)

Explanation: The program is not valid. Try replacing “int *p;” with “int *p = NULL;” and it will try to dereference a null pointer.
This is because fun() makes a copy of the pointer, so when malloc() is called, it is setting the copied pointer to the memory location, not p. p is pointing to random memory before and after the call to fun(), and when you dereference it, it will crash.
If you want to add memory to a pointer from a function, you need to pass the address of the pointer (ie. double pointer).

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