Exposing static pointer in some source file to main.c via function [duplicate]

Question

Pointer s doesn't get any value. I can't figure out why.
Variable v in exposeStatic() always get desired values.
Please advise.

main.c

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

struct node *s = NULL;

int main() 
{
    populateStatic();
    exposeStatic(s);
    printf("myLib myStatic adress: %p\n", s);
    return(0);
}

myLib.h

#ifndef myLib
#define myLib

struct node
{
    int data;
    struct node *next;
};

void exposeStatic(struct node *v);
void populateStatic();

#endif

myLib.c

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

static struct node *myStatic = NULL;

void exposeStatic(struct node *v)
{
    v = myStatic;
}

void populateStatic()
{
    struct node *p = (struct node *)malloc(sizeof(struct node));
    p->data = 855; // assume there is some data from populateStatic() arguments
    p->next = NULL; // assume there is some data from populateStatic() arguments
    myStatic = p;
}

v = myStatic; overwrites the value passed (if the compiler doesn't optimise it away), and is then forgotten about. — Weather Vane
– Weather Vane, Commented Feb 25, 2020 at 15:43
Does this answer your question? make the parameter as reference — RubberBee
– RubberBee, Commented Feb 25, 2020 at 15:45
Does this answer your question? Changing address contained by pointer using function — medalib
– medalib, Commented Feb 25, 2020 at 15:46

S.S. Anne · Accepted Answer · 2020-02-25 15:45:04Z

3

You're overwriting the local copy of the variable. Instead, pass a pointer to a pointer:

    exposeStatic(&s);

and modify your function accordingly:

void exposeStatic(struct node **v)
{
    *v = myStatic;
}

answered Feb 25, 2020 at 15:45

S.S. Anne

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