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From: "h2+bugs at fsfe dot org" <gcc-bugzilla@gcc.gnu.org>
To: gcc-bugs@gcc.gnu.org
Subject: [Bug c++/107104] semantics of __builtin_constant_p within
 static_assert and return value
Date: Mon, 03 Oct 2022 14:48:53 +0000
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https://gcc.gnu.org/bugzilla/show_bug.cgi?id=3D107104

Hannes Hauswedell <h2+bugs at fsfe dot org> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
                 CC|                            |h2+bugs at fsfe dot org

--- Comment #1 from Hannes Hauswedell <h2+bugs at fsfe dot org> ---
It seems that __builtin_constant_p does not indicate whether something *can=
 be*
a constant but whether *it is* a constant.
If you evaluate it in a non-const-context, the expression passed as argument
may or may not be evaluated at compile-time, so the value of
__builtin_constant_p may or may not be 1.

This example illustrates the behaviour:

```cpp
constexpr void foobar() {}    // =E2=86=92 test() returns 0 or 1
//consteval void foobar() {}  // =E2=86=92 test() returns 1

#define TEST_EXPR (foobar(), 0)

int test() {
    static_assert(__builtin_constant_p(TEST_EXPR));
    return __builtin_constant_p(TEST_EXPR);
}
```


To enforce evaluation in a const-context, you can define a macro like this:

#define IS_CONSTEXPR(...) std::integral_constant<bool,
__builtin_constant_p((__VA_ARGS__, 0))>::value=