Why is a type qualifier on a return type meaningless?

Question

Say I have this example:

char const * const foo( ){    /* which is initialized to const char * const */    return str; } 

What is the right way to do it to avoid the compiler warning "type qualifier on return type is meaningless"?

Answer 1

The way you wrote it, it was saying "the returned pointer value is const". But non-class type rvalues are not modifiable (inherited from C), and thus the Standard says non-class type rvalues are never const-qualified (right-most const was ignored even tho specified by you) since the const would be kinda redundant. One doesn't write it - example:

  int f();   int main() { f() = 0; } // error anyway!    // const redundant. returned expression still has type "int", even though the    // function-type of g remains "int const()" (potential confusion!)   int const g();  

Notice that for the type of "g", the const is significant, but for rvalue expressions generated from type int const the const is ignored. So the following is an error:

  int const f();   int f() { } // different return type but same parameters 

There is no way known to me you could observe the "const" other than getting at the type of "g" itself (and passing &f to a template and deduce its type, for example). Finally notice that "char const" and "const char" signify the same type. I recommend you to settle with one notion and using that throughout the code.