C | Dynamic Memory Allocation | Question 3

Output? 

C

# 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",*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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