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Subject: [Bug c++/55425] constexpr does not work in many situations  (both built-in and user supplied literals)
Date: Wed, 21 Nov 2012 12:13:00 -0000
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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=55425

--- Comment #1 from Jonathan Wakely <redi at gcc dot gnu.org> 2012-11-21 12:12:19 UTC ---
(In reply to comment #0)
> 
> A return statement is not a return statement if the returned value is __func__
> (also true for non-standard identifiers like __PRETTY_FUNCTION__).
> 
> // good
> //static const char func[] = "function-name";
> //constexpr const char* x() { return func; }
> 
> // bad
> constexpr const char* x() { return __func__;}
> 
> int main() { __builtin_puts(x()); return 0; }

The standard says __func__ is a function-local variable, defined as if by

constexpr const char* x() {
  static const char __func__[] = "function-name ";
  return __func__;
}

Clearly this is not a valid constexpr function.
Changing this would be an extension.


> Situation 2: user literals
> --------------------------
> 
> The (obviously constant) string that the compiler builds from the literal is
> not constant according to the compiler:
> 
> #include <stdio.h>
> 
> constexpr int valid_bin_number(const char* c) { return *c ? ((*c == '1' || *c
> == '0') ? valid_bin_number(c+1) : false ) : true; }
> 
> unsigned int operator"" _bin(const char* str)
> {
> static_assert(valid_bin_number(str), "not a binary number");

'str' is not a constant expression, so 'valid_bin_number(str)' is not a
constant expression either. This is not a bug.


> Situation 3: __m128i type
> --------------------------
> 
> Assigning a literal value to a constexpr __m128 fails because the literal is
> not a literal.

No, the error says __m128 is not a literal type, which I assume is true.
Changing that would be an enhancement request.