Question

I’m trying to pass a char* and change it in another function but somehow it keeps giving seg fault.

Full code is:

#include <stdio.h>

void getString(char** str) {
    *str[0] = '$';
    char c;
    int i = 1;
    while ((c = getchar()) != '$') {
        if (c != '\n') {
            *str[i-1] = c;
            i++;
        }
    }
    *str[i] = '\0';
}

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

    char* str =  (char*)malloc(200 * sizeof(char));
    while (1) {
        getString(&str);
        printf("String: %s\n",str);
    }
    return 0;
}

If I take the * from str[0] = ‘$’ it gives a warning passing pointer to integer.

Dunno where I’m messing up.

Answer 1

the array index , [], operator has lower precedence than the dereference operator, *

So you’d have to change your code to do

 (*str)[0] = '$';
 (*str)[i-1] = c;
 (*str)[i] = '\0';

In your case, you don’t need to pass in the address of your pointer, just pass the pointer:

getString(str);

And access it as

str[0] = '$';  
str[i-1] = c;
str[i] = '\0';

Since you’re using malloc(), be sure to add #include <stdlib.h>
If you didn’t cast the return value of malloc, which is not needed in C, you should have
gotten a warning regarding this.