c++ - 为什么编译器试图复制而不是 move 返回值？

Question

我的代码看起来像

// definition of class 'Foo'
class Foo
{
  private:
    vector<Bar> v;
  public:
    ...
    Foo(Foo&&) = default; // move constructor
};

// definition of function 'f'
Foo f()
{
  Foo x;
  DoStuff(x);
  return x;
}

// Somewhere in main
result = f(); // I am aiming to move 'x' to 'result'

当我尝试编译时收到

EAL_1.6.cpp:314:13: error: object of type 'Foo' cannot be assigned because its copy assignment operator is implicitly deleted
        x = f(x);
            ^
EAL_1.6.cpp:232:5: note: copy assignment operator is implicitly deleted because 'Foo' has a user-declared move constructor
    Foo(Foo&&) = default;
    ^

我很想试试

return move(x);

但根据 this post 这似乎不是一个聪明的解决方案.我知道在定义 move 构造函数时删除了 cop 构造函数(如 this post 中所述)，但我不明白如何告诉编译器我希望将“x” move 到“结果”。

Answer 1

这个:

result = f(); // I am aiming to move 'x' to 'result'

不是尝试 move 构造，而是尝试 move 分配。而且，正如编译器告诉您的那样:

object of type Foo cannot be assigned because its copy assignment operator is implicitly deleted

复制赋值运算符被隐式删除(并且 move 赋值运算符完全不存在)因为您添加了 move 构造函数。据推测，如果您的类型是可 move 构造的，那么它也是可 move 分配的。所以只需添加:

Foo& operator=(Foo&& ) = default;