change a var using pointer in a function

Question

#include <stdio.h>
#include <stdlib.h>

change (int i)
{
    int *x;
    x = &i;
    printf("%d\n",*x);
    *x = 7;
    printf("%d\n",*x);

} 

int main()
{
    int i=66 ; 
    change(i);
    printf("%d\n",i);
    return 0;
}

I'm wondering how to change a var value using pointer in a function using C.

For starters you'd need the function to actually accept a pointer as argument — UnholySheep, Oct 10 '22 at 15:42
Related: https://stackoverflow.com/questions/2229498/passing-by-reference-in-c — TylerH, Oct 11 '22 at 13:48
You could start by writing an actual valid function that compiles cleanly, then take it from there. — Lundin, Oct 11 '22 at 13:50

score 0 · Answer 1 · answered Oct 10 '22 at 15:44

0

You need to pass the variable i to the function by reference.

In C passing by reference means passing an object indirectly through a pointer to it. Thus dereferencing the pointer within the function you can change the original object pointed to by the pointer.

So your function will look like

void change (int *p)
{
    printf("%d\n",*p);
    *p = 7;
    printf("%d\n",*p);
}

and be called like

change( &i );

answered Oct 10 '22 at 15:44

Vlad from Moscow

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Surely, Vlad, you can find a duplicate for this. – Adrian Mole Oct 10 '22 at 21:43

change a var using pointer in a function

1 Answers1