pointers when passed as parameters

I've been wondering: what happens to a pointer when it's passed as a parameter to a function? Why is this:

<code>

int main(void) {

char *str;
*str = 'a';

return 0;

</code>

not legal, while this is perfectly legal:

<code>

void function(char *);

void function(char *str) {

*str = 'a';

}

int main(void) {

char *str;

function(str);

return 0;

}

</code>

Thanks
Skarr

I know that, but if I initialize the string in main() I would have a locked size for my string. And if i read stuff from (say) a file, I won't know how big tje buffer should be.

I don't think it's a good idea either way. Your char* points to garbage, so you need to point it to an actual buffer:

Code:

---

void funct(char* buffer, int size)
{
  /* do whatever operation you want on up to size
        bytes of the buffer */
}

int main()
{
    /* obviously you must define MAX_BUFFER_SIZE to
        something approprate somewhere */
    char buffer[MAX_BUFFER_SIZE];
    char* pbuffer = buffer;

    /* of course you could just pass buffer too */
    funct(pbuffer, MAX_BUFFER_SIZE);

    return 0;
}

---