inline consteval unsigned char operator""_UC(const unsigned long long n)
{
    return static_cast<unsigned char>(n);
}

inline consteval char f1(auto const octet)
{
    return char(octet >> 4_UC);
}

inline constexpr char f2(auto const octet)
{
    return char(octet >> 4_UC);
}

int main()
{
    auto c1 = f1('A'); // ok
    auto c2 = f2('A'); // error

    return c1   c2;
}

Compiled with: clang -14 -std=c 20 -stdlib=libc , and the error message is:

undefined reference to `operator"" _UC(unsigned long long)'

See online demo

Why does f1() is ok while f2() is not?

Why does consteval not behave as expected?

CodePudding user response：

They do behave as expected. The diagnostic could be better, but the behavior is certainly what one should expect.

auto c2 = f2('A'); // error

Since c2 is not a constexpr variable, its initialisation doesn't have to be constant evaluated. And it in fact isn't. This means the full-expression of its initialisation is not a constant expression, and so f2 is a "runtime" function here.

Meanwhile char(octet >> 4_UC) is an immediate invocation of operator""_UC. There's a "shall requirement" attached to it.

[expr.const] (emphasis mine)

13 An expression or conversion is in an immediate function context if it is potentially evaluated and its innermost non-block scope is a function parameter scope of an immediate function. An expression or conversion is an immediate invocation if it is a potentially-evaluated explicit or implicit invocation of an immediate function and is not in an immediate function context. An immediate invocation shall be a constant expression.

For c2, that requirement is violated. The code is ill-formed.