c++ - 为什么将 "&& true"添加到约束会使函数模板成为更好的重载？

Question

考虑以下两个函数模板的重载 foo :

template <typename T>
void foo(T) requires std::integral<T> {
    std::cout << "foo requires integral\n";
}

template <typename T>
int foo(T) requires std::integral<T> && true {
    std::cout << "foo requires integral and true\n";
    return 0;
}

请注意两个约束之间的区别:第二个约束有一个额外的 && true .
直观地说，true在连词中是多余的(因为 X && true 只是 X )。但是，看起来这在语义上有所不同，如 foo(42)会 call the second overload .
为什么会这样？具体来说，为什么第二个函数模板是更好的重载？

Answer 1

根据 [temp.constr.order] ，特别是 [temp.constr.order]/1和 [temp.constr.order]/3

/1 A constraint P subsumes a constraint Q if and only if, [...] [ Example: Let A and B be atomic constraints. The constraint A ∧ B subsumes A, but A does not subsume A ∧ B. The constraint A subsumes A ∨ B, but A ∨ B does not subsume A. Also note that every constraint subsumes itself. — end example ]

/3 A declaration D1 is at least as constrained as a declaration D2 if

• (3.1) D1 and D2 are both constrained declarations and D1's associated constraints subsume those of D2; or
• (3.2) D2 has no associated constraints.

如果我们考虑 A如 std::integral<T>和 B如 true ;然后:

A ∧ B这是 std::integral<T> && true包含 A , 即 std::integral<T> ,

这意味着对于以下声明:

// Denote D1
template <typename T>
void foo(T) requires std::integral<T> && true;

// Denote D2
template <typename T>
void foo(T) requires std::integral<T>;

D1 的相关约束包含 D2 的那些，因此 D1至少与 D2 一样受约束.同时反向不成立，D2至少不像 D1 那样受约束.这意味着，根据 [temp.constr.order]/4

A declaration D1 is more constrained than another declaration D2 when D1 is at least as constrained as D2, and D2 is not at least as constrained as D1.

声明D1比声明更受约束 D2 , 和 D1随后根据 [temp.func.order]/2 被重载解析选择为最佳匹配:

Partial ordering selects which of two function templates is more specialized than the other by transforming each template in turn (see next paragraph) and performing template argument deduction using the function type. The deduction process determines whether one of the templates is more specialized than the other. If so, the more specialized template is the one chosen by the partial ordering process. If both deductions succeed, the partial ordering selects the more constrained template (if one exists) as determined below.