Why I get; initializing 'char *' with an expression of type 'const char *' discards qualifiers?

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         


        

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).