c++ - 移动赋值运算符的异常说明符如何影响移动构造函数的异常说明符？

Question

我在 C++14 模式下使用 GCC 5.2 和 clang 3.6 进行测试，它们提供相同的输出。

对于下面的代码

#include <iostream>
#include <type_traits>

struct S {
  // S& operator= (S&&) noexcept { return *this; }
};


int main() {
  std::cout << std::is_nothrow_move_constructible<S>::value
            << std::is_nothrow_move_assignable<S>::value;  
}

得到结果11。但如果取消注释移动赋值运算符，输出将变为 01。移动赋值运算符的显式 noexcept 规范如何影响移动构造函数的规范？

Answer 1

通过定义移动赋值运算符，由于 rule of 5，您禁用了移动构造函数.该类不是 is_nothrow_move_constructible 因为它根本不可移动构造，除非您定义该构造函数，否则该构造函数不再可用。

§12.8 复制和移动类对象

If the definition of a class X does not explicitly declare a move constructor, one will be implicitly declared as defaulted if and only if
— X does not have a user-declared copy constructor,
— X does not have a user-declared copy assignment operator,
— X does not have a user-declared move assignment operator,
— X does not have a user-declared destructor, and
— the move constructor would not be implicitly defined as deleted.

在您没有用户定义的移动构造函数的情况下，两者都被隐式定义并遵循以下规范。

§15.4 异常规范

An implicitly declared special member function shall have an exception-specification. If f is an implicitly declared default constructor, copy constructor, move constructor, destructor, copy assignment operator, or move assignment operator, its implicit exception-specification specifies the type-id T if and only if T is allowed by the exception-specification of a function directly invoked by f’s implicit definition; f shall allow all exceptions if any function it directly invokes allows all exceptions, and f shall allow no exceptions if every function it directly invokes allows no exceptions.