没有调用移动控制器和复制器

Question

我们来看下面的C ++示例：

#include <iostream>

struct X
{
  std::string s;

  X() : s("X") { }

  X(const X& other) : s{other.s} { std::cout << "cpy-ctor\n"; }

  X(X&& o): s{o.s} { o.s = ""; std::cout << "move-ctor\n"; }

  X& operator=(const X& other) {
    std::cout << "cpy-assigned\n";
    s = other.s;
    return *this;
  }

  X& operator=(X&& other) {
    if (this != &other) {
      s = other.s;
      other.s = "";
    }
    std::cout << "move assigned\n";
    return *this;
  }
};

X f(X x) {
  std::cout << "f: ";
  return x;
}

X g() {
  std::cout << "g: ";
  X x;
  return x;
}

int main() {
  X x;
  X y;
  x = f(X());
  y = g();
}

如果我用gcc 4.8.2编译它，我得到以下结果：

f: move-ctor
move assigned
g: move assigned

我不明白为什么在调用g函数时没有调用copy-constructor。

我只是想了解何时调用复制或移动构造函数。

Answer 1

在C ++中，所有表达式都是：

• 左值
• prvalue
• x值

构建对象

- 左值

X x{};
Y y{x}; // copy constructor, x is an lvalue

- prvalue - RVO

使用RVO，默认情况下由gcc启用。这样就可以使用复制构造函数，而是构造一次对象。

X g()
{
  X x {}; 
  x.value = 10;
  return x;
}

X y {g()}; // X constructor get's called only once to create "y". Also 
           // y is passed a a reference to g() where y.value = 10. 
           // No copy/move constructor for optimization "as if" rule

- prvalue - 没有RVO

没有RVO，在这种情况下取​​决于它。如果显式或隐式删除了移动构造函数，那么它将调用复制构造函数

复制构造函数

struct X { X(const X&) {}}; // implicitly deletes move constructor 
                            // and move assignment, see rule of 5

X g()
{
  return X{}; // returns a prvalue
}

X y {g()}; // prvalue gets converted to xvalue, 
           // "temporary materialization", where the xvalue has an 
           // identity where members can be copied from. The xvalue
           // binds to lvalue reference, the one from copy constructor
           // argument

移动构造函数

X { X(X&&) {}}; // explicitly declared move constructor

X g()
{
  return X{}; // returns a prvalue
}

X y {g()}; // prvalue gets converted to xvalue, 
           // "temporary materialization", where the xvalue has an
           // identity where members can be moved from. The xvalue 
           // binds to rvalue reference, the one from move constructor 
           // argument

- xvalue

X x {};
X y {std::move(x)}; // std::move returns an xvalue, where if move 
                    // constructor is declared will call it, other wise 
                    // copy constructor, similar to explained above for 
                    // prvalue.

复制/移动作业

- 左值

X x{};
X y{};

x = y; // call copy assignment operator since y is an lvalue.

- prvalue

如果显式或隐式删除移动分配，则它将调用复制赋值运算符。

复制作业

struct X{ X& operator=(const X&); } // implicilty deletes move 
                                    // constructor and move assignment,
                                    // see rule of 5

X g()
{
  return X{}; // returns a prvalue
}

x = y; // prvalue gets converted to xvalue, 
       // "temporary materialization", where the xvalue has an identity 
       // where members can be copied from. The xvalue binds to lvalue 
       // reference, the one from copy assignment operator argument

移动作业

struct X{ X& operator=(X&&); } // explicitly declared move assignment        operator

X g()
{
  return X{}; // returns a prvalue
}

x = y; // prvalue gets converted to xvalue, 
       // "temporary materialization", where the xvalue has an identity 
       // where members can be moved from. The xvalue binds to rvalue 
       // reference, the one from move assignment operator argument

- xvalue

X x {};
X y {};

x = std::move(x); // std::move returns an xvalue, where if move
                  // assignment is declared will call it, other 
                  // wise copy assignment, similar to explained 
                  // above for prvalue. 

Answer 2

使用同一类型的另一个对象实例化对象时，将调用复制构造函数。

例如：

X x;
X y(x);

代码中的最后一行将函数返回的值赋给已构造的对象。这是通过移动分配完成的。