Question

I am trying to deliver a char pointer to a function which then fills it with stuff so that the calling function could use it then. As it always gave me weird stuff in the calling function i wrote a simple representation about what I have done:

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

void bla(char *t)
{
    t = (char*) malloc(5);
    if (t != 0) {
        t[0] = 'h';
        printf("%c\n", t[0]);
    }
}

int main(int argc, char **argv)
{
    char b[10];

    bla(b);
    printf("%c\n", b[0]);
    return 1;
}

I’m not quite sure if it has something to do that C passes a copy of the argument. Do I need to pass a pointer to a pointer then or is there a better solution?

EDIT:

Sorry guys but I didn’t get it. Could you please look over this example:

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

void blub(char **t)
{
    printf("%d\n", *t);
    *t = (char*) malloc(sizeof(char) * 5);
    *t[0] = 'a';
    *t[1] = 'b';
    printf("%d\n", *t);
    printf("%d\n", *t[0]);
    printf("%d\n", *t[1]);
}

int main(int argc, char **argv)
{
    char *a;
    blub(&a);
    printf("%d\n", a);
    printf("%d\n", a[0]);
    printf("%d\n", a[1]);

    return 1;
}

Output is as follows:

./main
6154128
140488712
97
98
140488712
97
0      <== THIS SHOULD BE 98 AS ABOVE!?

Why do i get 98 in the function blafu and in main it is a null pointer?! I am totaly confused :/

Answer 1

You need to use pointer to pointer.

//............vv
void bla(char **t)

Then use the pointer by dereferencing it:

// note the '*' in front of t
*t = (char*) malloc(5);
if (*t != 0) {
    *t[0] = 'h';
    printf("%c\n", *t[0]);
}

Also, declare b as char*, not as char b[10].

---

Why? Because you’re trying to change the pointer. The logic is the same as other types, but it’s kinda confusing here.
Think about it like this: if you need to pass an int and you need to change it, you need a pointer to int. The same is here – you need to pass a pointer to char and you need yo change it, so use “pointer to pointer to char” 🙂

---

EDIT:
According to your edit – yes, you have understood this perfectly, there’s just a small problem – the priority of operator* and operator[]. If you replace your *t[X] with (*t)[X], everythng will be fine 🙂