C programming

Question

#include<stdio.h>

void fun(int *p,int *q)
{
    p=q;
    *p=q;
}
int i=0,j=1;
int main()

{
    fun(&i,&j);
    printf("%d%d",i,j);
}

What will be the output of i and j in 16-bit C Compiler?

Comments

@Vikas Verma

*p=q;

*p means it is integer, q means integer pointer.... so, you can't assign them

---

On execution, I got a warning but not a compilation error for that.

---

did you run it on offline ? because online compilers does some error recovery...

But the concept is you can't assign them, even offline compilers also gives only warning message.

After that it is your wish.

Answer 1

#include<stdio.h>

void fun(int *p,int *q) 
{ 
    p=q;                    /*3. p was pointing to i but now content of p are changed to address of q */
    *p=q;                   /*4. p is pointing to j so now content of j are changed with its own address */
} 
int i=0,j=1;                /*1. global variables i and j defined */
int main()

{ 
   fun(&i,&j);             /*2. function fun is called and addresses of i and j are passed as parameters */
   printf("%d%d",i,j);     /*5. 0 address-of-j-in-int */
}

Output : 0 address-of-j-in-int

https://ideone.com/CMYOAh