5

按约束进行偏序的规则是指 AND 和 OR，但不指 NOT：

13.5.4 Partial ordering by constraints [temp.constr.order]
(1.2) ...
- 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.

这些规则基于原子约束和约束规范化的定义：

13.5.3 Constraint normalization [temp.constr.normal]
1 The normal form of an expression E is a constraint that is defined
  as follows:
(1.1) The normal form of an expression ( E ) is the normal form of E.
(1.2) The normal form of an expression E1 || E2 is the disjunction
      of the normal forms of E1 and E2.
(1.3) The normal form of an expression E1 && E2 is the conjunction
      of the normal forms of E1 and E2.

否定（即！E1）特别不处理。

因此，以下代码正确使用了部分排序：

void foo(auto i) requires std::integral<decltype(i)> {
    std::cout << "integral 1" << std::endl;
}

void foo(auto i) requires std::integral<decltype(i)> && true {
    std::cout << "integral 2" << std::endl;
}

int main() {
    foo(0); // integral 2
}

虽然此代码因模棱两可而失败：

template<typename T>
concept not_integral = !std::integral<T>;

template<typename T>
concept not_not_integral = !not_integral<T>;

void foo(auto i) requires not_not_integral<decltype(i)> {
    std::cout << "integral 1" << std::endl;
}

void foo(auto i) requires std::integral<decltype(i)> && true {
    std::cout << "integral 2" << std::endl;
}

int main() {
    foo(0);
}

代码：https ://godbolt.org/z/RYjqr2

以上导致德摩根定律不适用于概念：

template<class P>
concept has_field_moo_but_not_foo
     = has_field_moo<P> && !has_field_foo<P>;

不等于：

template<class P>
concept has_field_moo_but_not_foo
     = !(has_field_foo<P> || !has_field_moo<P>);

第一个将参与偏序，而后者则不会。

代码：https ://godbolt.org/z/aRhmyy

Was the decision, not to handle negation as part of constraint normalization, taken in order to ease the implementation for compiler vendors? or is there a logical flaw with trying to support it?

4

1 回答 1

7

Was the decision, not to handle negation as part of constraint normalization, taken in order to ease the implementation for compiler vendors?

Yes. This generalizes to requiring a SAT solver in the compiler.

There was an example added in [temp.constr.op]/5 to demonstrate this, although it does not provide the rationale for the decision:

template <class T> concept sad = false;

template <class T> int f1(T) requires (!sad<T>);
template <class T> int f1(T) requires (!sad<T>) && true;
int i1 = f1(42);        // ambiguous, !sad<T> atomic constraint expressions ([temp.constr.atomic])
                        // are not formed from the same expression

template <class T> concept not_sad = !sad<T>;
template <class T> int f2(T) requires not_sad<T>;
template <class T> int f2(T) requires not_sad<T> && true;
int i2 = f2(42);        // OK, !sad<T> atomic constraint expressions both come from not_­sad

template <class T> int f3(T) requires (!sad<typename T::type>);
int i3 = f3(42);        // error: associated constraints not satisfied due to substitution failure

template <class T> concept sad_nested_type = sad<typename T::type>;
template <class T> int f4(T) requires (!sad_nested_type<T>);
int i4 = f4(42);        // OK, substitution failure contained within sad_­nested_­type

In particular, note the difference between f3 and f4. Does requires !sad<typename T::type> mean that there be no sad nested type, or that there be a nested type that is not sad? It actually means the latter, while the constraint on f4 means the former.

于 2020-02-28T14:38:40.960 回答