pointers

Question

void fun(int *p)
{
  int q = 10;
  p = &q;
  printf("%d",p);//prints 1948738932
  printf("\n");
  printf("%d q= ",&q);//prints 1948738932
  printf("\n");
  printf(" value of p  %d" ,*p);//rints 10
  printf("\n");
}

int main()
{
  int r = 20;
  int *p = &r;
  printf("%d r= ",&r);//output = 1948738964
  printf("\n");
  printf("%d",p);//output = 1948738964
  printf(" value of p  %d" ,*p);// prints 20
  printf("\n");
  fun(p);
  printf("%d", *p);//prints 20
  return 0;
}

Can anyone explain why is it giving 20 in line 2 because in function fun, address of p has been changed to point to q,so it should print value at q i.e 10??

In line 3,what is passed in the function?is it the address of p or something else?

Comments

It will print 20,10,20
because p of main() and p of fun() are different
p of main is local to main()
So, p is passing address of value 20

then in fun()
local pointer p  pointing to address of q
So, there is no change in main()

Answer 1

Assume address of r = 1000

So, in the main function is p is holding the value = 1000

Now you called fun(p);

What, happens here ?

New stack frame is allocated for funciton fun().

Argument of fun() is int *p so a new pointer variable named p is also created in this stack frame and p is initialized to 1000 because of parameter passing. So far so good ! :)

Now, again another local varibale is q is allocated and initialized to 10. Assume its adress is 2000.

Then, inside fun() value of local pointer varibale p is changed to 2000.

Keep in mind that p in the stack frame of fun() is different from p in the main function. In the main funciton p is pointing to 1000.

after that you have done some printing stuff inside the fun().

Did you change pointer p in the main function ? NO

Did you change the value contained in the address 1000 ? NO.

So, why should the *p or r value change in the main function ? No change.

Comments

let me denote p in main function by P just to distinguish between two p's.

P points to r, thus it stores the address of r in P. And when it is passed to function fun(P), In the called function declaration is  fun(int *p), p in fun() is a pointer to integer, so it takes the integer value stored in P, that is the address of r and continues with it. p is a local variable in fun(), so after it is executed everything about the function fun() is removed from stack frame. and when control reaches back to main function P here is a pointer of interge, that still points to r = 20.

Watch this : https://www.youtube.com/playlist?list=PL2_aWCzGMAwLZp6LMUKI3cc7pgGsasm2_

---

i got it now..thankyou so much

---

thanks for the xplanation and the link

Answer 2

in  fun(p){p is holding the local address of q so prints 10}

on returning to main p still holds the address of r so prints 20