“ constexpr”变量“用于其自身的初始化程序”：Clang vs. GCC [英] `constexpr` variable "used in its own initializer": Clang vs. GCC

问题描述

这个问题似乎与现有的问题有关，但是我不理解答案（涉及 const auto this_ = this; ），而且我认为以下示例更容易理解。

This question seems related to an existing one, but I do not understand the "portable workaround" provided in the answer there (involving const auto this_ = this;) and moreover I think the following example is easier to follow.

我正在使用以下C ++ 17代码段（实时演示）：

I am playing with the following snippet of C++17 code (live demo):

#include <iostream>

struct Test {
  const char* name_{nullptr};
  const Test* src_{nullptr};

  constexpr Test(const char* name) noexcept
    : name_{name}
  {}

  constexpr Test(const Test& src) noexcept
    : src_{&src}
  {
    name_ = src_->name_;
    src_ = nullptr;
  }
};

template<char c>
void print_constexpr_char() {
    std::cout << c << std::endl;
}

int main() {
  constexpr const char* in = "x";
  constexpr auto foo = Test{in};
  constexpr auto bar = Test{foo};
  std::cout << bar.name_ << std::endl;

  print_constexpr_char<bar.name_[0]>();

  return 0;
}

在GCC 7.2中编译失败，而Clang 5.0.0则看不到任何问题。 GCC错误本质上是

Compilation fails with GCC 7.2 while Clang 5.0.0 does not see any problem. The GCC error essentially reads

错误：'bar'的值不能在常量表达式中使用

error: the value of 'bar' is not usable in a constant expression

注意：在自己的初始化程序中使用的'bar'

note: 'bar' used in its own initializer

在意识到删除了final print_constexpr_char 使代码编译，尽管它仍包含GCC使用的行 constexpr auto bar = Test {foo}; 抱怨（用在它自己的初始化程序中）。

I am even more confused after realizing that removing the final print_constexpr_char makes the code compile although it still contains the line constexpr auto bar = Test{foo}; which GCC used to complain about ("used in its own initializer").

1. 哪个编译器在这里正确？


2. 如何理解GCC说明（如果不是bug），如果随后在常量表达式中使用结果，则在其自身的初始化程序中使用是有害的吗？


3. 是否存在有效的在 constexpr 构造函数中使用指针作为中间阶段的方式/解决方法，然后将正在构造的对象转换为可以存储在 constexpr中的最终状态变量？

1. Which compiler is correct here?
2. How to understand the GCC note (if not a bug) that "using in its own initializer" is harmful iff the result is subsequently used in a constant expression?
3. Is there a valid way/workaround to use pointers in a constexpr constructor as an intermediate stage before transforming the object under construction into the final state which can be stored in a constexpr variable?

推荐答案

GCC拒绝代码是正确的（但是错误消息可能会起作用）。除非该变量具有静态存储持续时间，否则无法在常量表达式中使用该变量的地址。

GCC is correct to reject the code (however the error message could use some work). You cannot use the address of a variable in a constant expression unless that variable has static storage duration.

foo 为不是静态的。如果将其移至 main 之外，则一切正常。 演示

foo is not static. If you move it outside of main, things will work. Demo

下面标记的行是问题所在：

The line marked below is the problem:

constexpr Test(const Test& src) noexcept
: src_{&src} <--- That

标准参考：（强调我的）

[expr.const]

[expr.const]

一个常量表达式要么是glvalue核心常量表达式，指的是一个实体，该实体是一个常量表达式（定义如下）的允许的
结果，或者是一个值满足以下条件的
的prvalue核心常量表达式：




（5.2）-如果值是指针类型，则它包含具有静态存储持续时间的对象的地址，即
超出此类对象末尾的地址（8.7），函数的地址或空指针值，

A constant expression is either a glvalue core constant expression that refers to an entity that is a permitted result of a constant expression (as defined below), or a prvalue core constant expression whose value satisfies the following constraints:

(5.2) — if the value is of pointer type, it contains the address of an object with static storage duration, the address past the end of such an object (8.7), the address of a function, or a null pointer value,

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