c++ - 为什么在我的代码中调用复制构造函数而不是移动构造函数？

Question

我试图理解移动构造函数和右值引用。所以我在 https://www.onlinegdb.com/online_c++_compiler 上尝试了这段代码.但结果让我感到困惑。

#include <iostream>
#include <type_traits>


class A {
public:
  A() { std::cout << "Constructed" << std::endl; }
  A(const A& )= delete;
  A(A&&) { std::cout << "Move Constructed" << std::endl; }
};

int
main ()
{
  A&& a = A();
  A b = a; // error: use of deleted function ‘A::A(const A&)’
  //A b = static_cast<decltype(a)>(a); // This works, WTF?
  std::cout << std::is_rvalue_reference<decltype(a)>::value << std::endl; // pretty sure a is rvalue reference.

  return 0;
}

Answer 1

您混淆了类型 和value categories .

(强调我的)

Each C++ expression (an operator with its operands, a literal, a variable name, etc.) is characterized by two independent properties: a type and a value category.

作为命名变量，a是一个左值。

The following expressions are lvalue expressions:

• the name of a variable, ...
• ...

然后为 A b = a;复制构造函数被选中。如您所试，static_cast<decltype(a)>(a);会将其转换为 xvalue(右值)；你也可以使用 std::move .

A b = std::move(a);

The following expressions are xvalue expressions:

• a function call or an overloaded operator expression, whose return type is rvalue reference to object, such as std::move(x);
• ...