有没有办法将参数转发给内部constexpr函数？ [英] Is there a way to forward argument to inner constexpr function?

问题描述

问题：是否可以通过将其参数传递给内部constexpr函数来对函数内部的常量表达式求值？
示例：

The question: is it possible to evaluate constant expression inside a function by passing (maybe with some kind of "perfect forwarding") its argument to inner constexpr function? Example:

constexpr size_t foo(char const* string_literal) {
    return /*some valid recursive black magic*/;
}

void bar(char const* string_literal) {
    // works fine
    constexpr auto a = foo("Definitely string literal.");
    // compile error: "string_literal" is not a constant expression
    constexpr auto b = foo(string_literal);
}

template<typename T>
void baz(T&& string_literal) {
    // doesn't compile as well with the same error
    constexpr auto b = foo(std::forward<T>(string_literal));
}

int main() {
    // gonna do this, wont compile due to errors mentioned above
    bar("Definitely string literal too!");
}

在文档，但找不到解决方案，也无法证明。内部表达式的构造性很重要。

Can't find anything clearly prohibiting in the documentation, but the solution isn't found, as well as a proof of impossibility. Constexpr'ness of inner expression is important.

推荐答案

constexpr 的参数在 constexpr 函数中不能将函数假定为 constexpr ；如果不是 constexpr ，则该函数必须正常工作。

Parameters to constexpr functions cannot be assumed to be constexpr within a constexpr function; the function must work if they are not constexpr.

类型参数可以。

如果您将 bar（ hello）替换为 bar（string_constant<'h '，'e'，'l'，'l'，'o'> {}）与 template< char ...> struct string_constant {}; ，现在将字符值编码在类型中，并且在路径中可用。还有其他方法可以将字符转换为类型。

If you replaced bar("hello") with bar( string_constant<'h', 'e', 'l', 'l', 'o'>{} ) with template<char...>struct string_constant{};, the value of the characters is now encoded in the type and will be available down the path. There are other ways to get the characters into a type.

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