Passing pointer argument by reference under C?

Question

#include 
#include 

void
getstr(char *&retstr)
{
 char *tmp = (char *)malloc(25);
 strcpy(tmp, \"hello,world\");
 retstr = tmp;
}         


        

Answer 1

References are a feature of C++, while C supports only pointers. To have your function modify the value of the given pointer, pass pointer to the pointer:

void getstr(char ** retstr)
{
    char *tmp = (char *)malloc(25);
    strcpy(tmp, "hello,world");
    *retstr = tmp;
}

int main(void)
{
    char *retstr;

    getstr(&retstr);
    printf("%s\n", retstr);

    // Don't forget to free the malloc'd memory
    free(retstr);

    return 0;
}