为什么 std::totally_ordered<float> return 是真的？

Question

cpp 参考资料 (https://en.cppreference.com/w/cpp/concepts/totally_ordered) 说 std::totally_ordered<T> 仅在给定 const std::remove_reference_t<T> 类型的左值 a、b 和 c 时建模：

恰好 bool(a < b)、bool(a > b) 和 bool(a == b) 之一为真；
如果bool(a < b)和bool(b < c)都为真，则bool(a < c)为真；
bool(a > b) == bool(b < a)
bool(a >= b) == !bool(a < b)
bool(a <= b) == !bool(b < a)

于是我考虑了一下NaN，发现float与bool(a > b) == bool(b < a)这句话不相符。但是 std::totally_ordered<float> 是 true。我做错了什么吗？

=======

我用这个宏来创建NaN,

#define NAN        ((float)(INFINITY * 0.0F))

这是我的代码：

#include <iostream>
#include <concepts>

using namespace std;

int main(int argc, char* argv[])
{
    /*
    1) std::totally_ordered<T> is modeled only if, given lvalues a, b and c of type const std::remove_reference_t<T>:
    Exactly one of bool(a < b), bool(a > b) and bool(a == b) is true;
    If bool(a < b) and bool(b < c) are both true, then bool(a < c) is true;
    bool(a > b) == bool(b < a)
    bool(a >= b) == !bool(a < b)
    bool(a <= b) == !bool(b < a)
    */
    constexpr bool b = totally_ordered<float>; // true
    cout << typeid(NAN).name() << endl;        // float
    cout << NAN << endl;
    cout << b << endl;

    cout << "Exactly one of bool(a < b), bool(a > b) and bool(a == b) is true;" << endl;
    cout << (NAN < NAN) << endl;
    cout << (NAN > NAN) << endl;
    cout << (NAN == NAN) << endl;

    cout << " If bool(a < b) and bool(b < c) are both true, then bool(a < c) is true;" << endl;
    cout << (1.f < 2.f) << endl;
    cout << (2.f < NAN) << endl;
    cout << (1.f < NAN) << endl;

    cout << "bool(a > b) == bool(b < a)" << endl; ////// IT IS FALSE //////
    cout << (NAN > 1.f) << endl;
    cout << (1.f < NAN) << endl;

    cout << "bool(a >= b) == !bool(a < b)" << endl;
    cout << (NAN >= 1.f) << endl;
    cout << (NAN < 1.f) << endl;

    cout << "bool(a <= b) == !bool(b < a)" << endl;
    cout << (NAN <= 1.f) << endl;
    cout << (NAN > 1.f) << endl;
    cout << endl;
}

Answer 1

概念有句法要求，存在一些表达式集并且是提供特定行为的类型。 C++20的concept特性可以检测到这些。

概念也有语义要求，关于表达的意义的要求，可能是相对的。 concept 功能不能（通常）检测到这些。如果一种类型满足句法和语义要求，则称其为概念“建模”。

对于totally_ordered，float满足概念的句法要求，但是IEEE754浮点数不满足语义要求。实际上，C++20 在符号中使用 totally_ordered<float> 作为 an example of this syntactic vs. semantic divide。

一些 concept 试图通过要求用户明确选择加入语义要求来解决这个问题。但是 totally_ordered 不是其中之一。

为什么 std::totally_ordered<float> return 是真的？

Why does std::totally_ordered<float> return true?

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