约束函数会允许模板参数依赖于函数参数吗? [英] Will consteval functions allow template parameters dependent on function arguments?

问题描述

在C ++ 17中，此代码是非法的:

In C++17, this code is illegal:

constexpr int foo(int i) {
    return std::integral_constant<int, i>::value;
}

那是因为即使foo可以在编译时求值，编译器仍然需要产生指令以在运行时执行它，从而使模板实例化成为不可能.

That's because even if foo can be evaluated at compile-time, the compiler still needs to produce the instructions to execute it at runtime, thus making the template instantiation impossible.

在C ++ 20中，我们将具有consteval函数，这些函数需要在编译时进行评估，因此应删除运行时约束.这是否意味着该代码是合法的?

In C++20 we will have consteval functions, which are required to be evaluated at compile-time, so the runtime constraint should be removed. Does it mean this code will be legal?

consteval int foo(int i) {
    return std::integral_constant<int, i>::value;
}

推荐答案

否.

无论本文将进行何种更改，这点上，它都无法更改以下事实: -template函数定义只键入一次.此外，如果您建议的代码合法，那么我们大概可以找到一种声明类型为std::integral_constant<int, i>的变量的方法，就ODR而言，这是非常禁止的.

Whatever changes the paper will entail, which is little at this point, it cannot change the fact that a non-template function definition is only typed once. Moreover, if your proposed code would be legal, we could presumably find a way to declare a variable of type std::integral_constant<int, i>, which feels very prohibitive in terms of the ODR.

本文还指出，在其示例之一中，参数不打算被视为核心常量表达式；

The paper also indicates that parameters are not intended to be treated as core constant expressions in one of its examples;

consteval int sqrsqr(int n) {
  return sqr(sqr(n)); // Not a constant-expression at this  point,
}                     // but that's okay.

简而言之，由于可能存在键入差异，因此函数参数永远不会是常量表达式.

In short, function parameters will never be constant expressions, due to possible typing discrepancy.

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