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[Bug c++/55425] constexpr does not work in many situations (both built-in and user supplied literals)
- From: "jakub at gcc dot gnu.org" <gcc-bugzilla at gcc dot gnu dot org>
- To: gcc-bugs at gcc dot gnu dot org
- Date: Wed, 21 Nov 2012 12:40:29 +0000
- Subject: [Bug c++/55425] constexpr does not work in many situations (both built-in and user supplied literals)
- Auto-submitted: auto-generated
- References: <bug-55425-4@http.gcc.gnu.org/bugzilla/>
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=55425
--- Comment #3 from Jakub Jelinek <jakub at gcc dot gnu.org> 2012-11-21 12:40:29 UTC ---
1) is not valid constexpr IMHO, as the standard says that __func__ works as if
static const char __func__[] = "function-name";
has been provided, but when it is provided, it is not valid constexpr.
2) I don't see why you think you could use something like this at all.
You are defining a non-constexpr operator "", so str there definitely isn't
pointer to a string literal, it is a standalone function that might be
inlined if the inliner choses so. So it is obvious the static_assert
which must be evaluated before the optimizations, without depending on
whether the function is inlined or not, must fail.
You want something like:
constexpr int
cstrlen (const char *c)
{
return *c ? cstrlen (c + 1) + 1 : 0;
}
constexpr unsigned int
bin_number (const char *c)
{
return *c ? (*c == '1' ? (1U << cstrlen (c + 1)) : *c == '0' ? 0 : throw 0) +
bin_number (c + 1) : 0;
}
constexpr unsigned int
operator"" _bin (const char *c)
{
return bin_number (c);
}
int
main ()
{
constexpr unsigned int a = 10000_bin;
(void) a;
return 0;
}
instead, then you can verify it the same at compile time.