Allocating memory to a struct member pointer in C

Question

I have a structure with a member that I need to pass to a function by reference. In that function, I'd like to allocate memory & assign a value. I'm having issues somewhere along the line - it seems that after the code returns from allocateMemory , the memory that I had allocated & the values that I assigned go out of scope (this may not be exactly what is happening, but it appears to be the case).

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

typedef struct myStruct_t
{
    char *myString;
} myStruct;

void allocateMemory(void *str);

int main(void) {
    myStruct tmp = {
        .myString = NULL
    };
    myStruct *p = &tmp;

    allocateMemory(p->myString);
    //allocateMemory(&(p->myString)); //also tried this

    printf("%s", p->myString);

    return 0;
}

void allocateMemory(void *str)
{
    str = malloc(8);

    ((char *)str)[0] = 'a';
    ((char *)str)[1] = 0;
}

If I print the value of str inside of allocateMemory , the 'a' is successfully printed, but if I attempt to print p->myString in main , my string is empty.

Can anyone tell me what I'm doing wrong?

Answer 1

You need to pass address of the structure member and then you can change (aka allocate memory) to it. In your version of the function, you are not taking a pointer not reference of a pointer, so you can change the content of memory referenced by the pointer but not the pointer itself.

So change your function to

void allocateMemory(char **ret_str)
{
    char *str = malloc(8);

    str[0] = 'a';
    str[1] = 0;
    *ret_str = str;
}

And then call it as

allocateMemory(&p->myString)

Answer 2

An alternative way of writing the same function Rohan did, eliminating the need to define any new variables:

void allocateMemory(char **str, size_t size) {
    *str = malloc(size);
    (*str)[0] = 'a';
    (*str)[1] = '\0';
}

Note that I pass a size parameter to justify using malloc() in the first place.