Passing a pointer to multiple functions and memory allocation in first function

Question

My question is in the context of C programming and NOT C++! I am trying to pass a pointer between multiple function. However the memory allocation should not be done by the caller. I tried a small example to simulate this. As can be seen, when the pointer points to a struct variable defined in main function, it is 'working' as expected. That is my function can manipulate the value in that memory address when the value of the address is passed. However when the function call returns and control passes on to main, why does the pointer get 'reinitialized'? Can the pointer somehow reflect the address that it was pointing to?

How can this be done?

Here is what I have:

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

//example to pass a struct to void pointer and a void pointer to struct
//testing if memory allocation is done by callee and not caller

typedef struct mystructure{
    int a;
    char b;
    unsigned char c[10];
}mystruct;


void func(void *var){

    mystruct *s = var;
    s->a = 100;
    s->b = 'I';
    strncpy(s->c,"test",5);
    printf("In func\n");
    printf("s->a = %d\n",s->a);
    printf("s->b = %c\n",s->b);
    printf("s->c = %s\n",s->c);
}

void voidOut(void *var){

    mystruct *s = var;
    printf("In voidOut\n");
    printf("s->a = %d\n",s->a);
    printf("s->b = %c\n",s->b);
    printf("s->c = %s\n",s->c);

}

//here is void pointer is both and 'in' and 'out' parameter
void memfunc(void *var){

    mystruct *s = var;
    s = (mystruct *)malloc(sizeof(mystruct));
    s->a = 100;
    s->b = 'I';
    printf("In memfunc\n");
    strncpy(s->c,"test",5);
    printf("s->a = %d\n",s->a);
    printf("s->b = %c\n",s->b);
    printf("s->c = %s\n",s->c);
}

//here is void pointer is an 'in' parameter
void memvoidOut(void *var){

    mystruct *s = var;
    printf("In memvoidOut\n");
    printf("s->a = %d\n",s->a);
    printf("s->b = %c\n",s->b);
    printf("s->c = %s\n",s->c);
}


int main(int argc, char *argv[]){
    mystruct val;
    func(&val);
    voidOut(&val);

    mystruct *ptr = NULL;
    memfunc(ptr);
    memvoidOut(ptr);

    return 0;
}

UPDATE: Following the answers and comments, here is what I have:

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

//example to pass a struct to void pointer and a void pointer to struct
//testing if allocation is done by callee and not caller

typedef struct mystructure{
    int a;
    char b;
    unsigned char c[10];
}mystruct;


void func(void *var){

    mystruct *s = var;
    s->a = 100;
    s->b = 'I';
    strncpy(s->c,"test",5);
    printf("In func\n");
    printf("s->a = %d\n",s->a);
    printf("s->b = %c\n",s->b);
    printf("s->c = %s\n",s->c);
}

void voidOut(void *var){

    mystruct *s = var;
    printf("In voidOut\n");
    printf("s->a = %d\n",s->a);
    printf("s->b = %c\n",s->b);
    printf("s->c = %s\n",s->c);

}


//here is void pointer is both and 'in' and 'out' parameter
void memfunc(void **var){

    mystruct *s = var;
    s = (mystruct *)malloc(sizeof(mystruct));
    s->a = 100;
    s->b = 'I';
    printf("In memfunc\n");
    strncpy(s->c,"test",5);
    printf("s->a = %d\n",s->a);
    printf("s->b = %c\n",s->b);
    printf("s->c = %s\n",s->c);
    //memcpy(var,s, sizeof(s));
}

//here is void pointer is an 'in' parameter
void memvoidOut(void **var){

    mystruct *s = var;
    printf("In memvoidOut\n");
    printf("s->a = %d\n",s->a);
    printf("s->b = %c\n",s->b);
    printf("s->c = %s\n",s->c);
}


int main(int argc, char *argv[]){
    mystruct val;
    func(&val);
    voidOut(&val);

    mystruct *ptr = NULL;
    memfunc(&ptr);
    memvoidOut(&ptr);

    return 0;
}

However my output is:

In func
s->a = 100
s->b = I
s->c = test
In voidOut
s->a = 100
s->b = I
s->c = test
In memfunc
s->a = 100
s->b = I
s->c = test
In memvoidOut
s->a = 0
s->b = d
s->c =

What am I missing? Shoud I define memory for the struct in memvoidOut?

When I explicitly do I memcpy(var,s,sizeof(s)) in memfunc, it works. However I would like to understand what is really happening — user489152
– user489152, Commented Jan 21, 2016 at 9:47
Why not change void memfunc(void *var) to void *memFunc()....return s; — Ed Heal
– Ed Heal, Commented Jan 21, 2016 at 9:50
Actually I should later use integers as return values to return just error codes.... — user489152
– user489152, Commented Jan 21, 2016 at 9:51

sestus · Accepted Answer · 2016-01-21 09:50:58Z

4

You should change the signature of the function that allocates the new memory:

void memfunc(void *var)

to

void memfunc(void **var)

and call from main

memfunc(&ptr)

Explanation:

Your memfunc function, makes a copy of the passed pointer ptr. Initially, it assigns this pointer to the allocated space (via malloc), but then, after the function returns, the pointer is still the original one that was passed to memfunc (that is NULL) because that ptr pointer's copy was reclaimed.

Now if you change it as proposed, the address of the pointer will be passed, thus you will not face that "copy pointer" issue

edited Jan 21, 2016 at 9:50

answered Jan 21, 2016 at 9:49

sestus

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9 Comments

user489152 Over a year ago

Thanks but I just added a comment: When I explicitly do I memcpy(var,s,sizeof(s)) in memfunc, it works

sestus Over a year ago

memcpy has nothing to do with your function. It is a tottaly different approach

user489152 Over a year ago

Oh I see. thanks for pointing out that the pointer passed is a copy!! That explains

sestus Over a year ago

Np, happy to help :)

sestus Over a year ago

In parameter meaning that you want to pass something other than null? And make some calculations before assigning new memory to it? Yes that should be doable too.

|

Sourav Ghosh · Accepted Answer · 2016-01-21 09:53:59Z

1

Function arguments are passed by value in C. So, you cannot pass the pointer itself and allocate memory to the pointer itself from the called function.

You can do either of two things

Pass the address of the pointer to allocated memory from another function.
Pass the pointer, allocate memory, return the allocated pointer from the function and in the caller, collect the return value in the same pointer.

So, in case of memfunc(), either

You have to change the signature to pass &ptr, collect in mystruct **var and then, inside the function, allocate memory for *var
You need to return a mystruct * and collect that in ptr.

edited Jan 21, 2016 at 9:53

answered Jan 21, 2016 at 9:47

Sourav Ghosh

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2 Comments

user489152 Over a year ago

In void memfunc(void *var) I was hoping to somehow keep the var parameter as both the 'input' and 'output' parameter

user489152 Over a year ago

Sourav Ghosh, could you please explain the second method. Can I keep the returned structure in the void pointer itself. Can you please elaborate

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