Question

I can’t figure out why I get this warning from clang by myself:

function_prototype_const_modifier.c:13:8: warning: initializing 'char *' with an
      expression of type 'const char *' discards qualifiers
      [-Wincompatible-pointer-types]
        char *ptr1 = source;
              ^      ~~~~~~
1 warning generated.

The code is very simple

#include<stdio.h>

char *my_strcpy(char *destination, const char *source);

int main(void) {
    char str1[] = "this is something";  
    char str2[] = "123456789123456789";
    my_strcpy(str2, str1);
    puts(str2);
    return 0;
}
char *my_strcpy(char *destination, const char *source) {
    char *ptr1 = source;
    char *ptr2 = destination;
    while(*ptr1 != '\0') {
        *ptr2++ = *ptr1++;
    }
    *ptr2 = '\0';
    return destination;
}

any idea?

Answer 1

source is a const char *, a pointer to const characters, so the characters cannot be changed by dereferencing the pointer (i. e. source[0] = 'A'; is a constraint violation).

However, assigning it to a char * discards this constraint; a simple char * suggests that the characters pointed to by the ptr1 pointer are not constant and you can now freely write ptr1[0] = 'A'; without getting compiler errors (a “diagnostic message”).

Consider what this means when you pass in a string literal. Since a string literal is “readonly” (it’s a const char []), trying to modify its contents is undefined behavior. So if you call

my_strcpy(destination, "Constant String");

but in the code for some reason you write

ptr1[0] = 'A';

you won’t get a compiler diagnostic message because ptr1 is a pointer to non-const chars, but your program will still invoke undefined behavior (and in practice, most likely crash, since string literals are placed in readonly memory regions).