Passing an char array as argument to a Callback in C

Question

How do we pass an char array as an argument to a callback? When I try to code something like:

void caller(void (*function)(char*), char* string)
{
     function(string);
}

void function(char* string)
{
     string[0]=data0;
     string[1]=data1;
     string[2]=data2;
}

void principal()
{
     char* str={0};
     caller(&function, str);
}

in the debugger watchlist, the string argument behaves like a common pointer (not an array) and the char array does not get populated with data at all. The function "principal" gets called in the main() when certain conditions are met.

Welcome to Stack Overflow, take a look at [How do I ask a good question?](https://stackoverflow.com/help/how-to-ask) — Yeheshuah, Jun 01 '21 at 01:37
Also possible duplicate. https://stackoverflow.com/questions/6567742/passing-an-array-as-an-argument-to-a-function-in-c — Yeheshuah, Jun 01 '21 at 01:39
Welcome to SO. What is this supposed to do? `char* str={0};` Don't you want to provide some memory for our callback function? Also: Your string argument **is** a pointer. Even if it was an array in the parameter list, it would decay to a plain pointer. — Gerhardh, Jun 01 '21 at 05:38
Which "char array" are you talking about? There is no array in your code. — Gerhardh, Jun 01 '21 at 05:40
Does this answer your question? [Passing an array as an argument to a function in C](https://stackoverflow.com/questions/6567742/passing-an-array-as-an-argument-to-a-function-in-c) — the busybee, Jun 01 '21 at 06:49

score 0 · Answer 1 · answered Jun 01 '21 at 07:34

You have not allocated any memory for str in principal() and your code have a big vulnerability about buffer overflow.

So I fixed your code by changing how str is declared and to add a size argument to function() and check it so that there is no overflow possible. For the rest, I kept the code as close as possible to what you showed:

char data0 = 'A';
char data1 = 'B';
char data2 = 'C';

void caller(void (*function)(char*, int size), char* string, int size)
{
    function(string, size);
}

void function(char* string, int size)
{
    if (size >= 3) {
        string[0] = data0;
        string[1] = data1;
        string[2] = data2;
    }
}

void principal()
{
    char str[] = { 0, 0, 0 };
    caller(&function, str, sizeof(str));
}

Passing an char array as argument to a Callback in C

1 Answers1