std::is_constant_evaluate 时如何获取常量表达式？

Question

Let's start with the example ( godbolt ):

constexpr int len(int v) {
    if (std::is_constant_evaluated()) {
        // static_assert(v > 0); // ERR: 'v' is not a constant expression
        return v;
    } else {
        return 0;
    }
}

using A = std::array<int, len(3)>;

The problem is, that the static_assert won't compile (gcc / clang latest 10.x releases). Apparently, v isn't realized to be constexpr when std::is_constant_evaluated returns true . But clearly, by the use of len , it actually is.

Question : Is it possible to use a variable as constexpr if and only if std::is_constant_evaluated ? If so, how?

Answer 1

Apparently, v isn't realized to be constexpr when std::is_constant_evaluated returns true . But clearly, by the use of len , it actually is.

No, it's actually not.

Function parameters are not constant expressions. It doesn't matter if you're in the middle of constant evaluation or not. You cannot use v as a constant expression in any context. static_assert requires a constant expression - this is why you cannot use v .

Question: Is it possible to use a variable as constexpr if and only if std::is_constant_evaluated ? If so, how?

No, it is not. Because in order to use a variable as a constant expression, well, that's a template. And is_constant_evaluated can't conditionally make your function into a function template, that's just not how the compilation process can work. See P0992 .

Answer 2

Not possible, but also unnecessary.

Instead of using a static_assert , you can throw something. Since you're in a constexpr context, it will give you a compile-time error.