Question

How can I make the following work? The idea is for the function to allocate an external pointer so I can use this concept in another program, but I can’t do that because gcc keeps telling me that the argument is from an incompatible pointer type… It should be simple, but I’m not seeing it.

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

int allocMyPtr(char *textToCopy, char **returnPtr) {
    char *ptr=NULL;
    int size=strlen(textToCopy)+1;
    int count;

    ptr=malloc(sizeof(char)*(size));
    if(NULL!=ptr) {
        for(count=0;count<size;count++) {
            ptr[count]=textToCopy[count];
        }
        *returnPtr = ptr;
        return 1;
    } else {
        return 0;
    }
}

int main(void) {
    char text[]="Hello World\n";
    char *string;

    if(allocMyPtr(text,string)) {
        strcpy(string,text);
        printf(string);
    } else {
        printf("out of memory\n");
        return EXIT_FAILURE;
   }
   free(string);
   return EXIT_SUCCESS;
}

Answer 1

It’s almost correct, but as your function wants a pointer to a pointer, you have to pass the address of the pointer to the function, using the address-of operator:

allocMyPtr(text, &string)