为什么移动构造函数的成员需要默认构造函数？ [英] Why does a move constructor require a default constructor for its members?

问题描述

我试图为没有复制构造函数的类实现move构造函数。我收到一个错误，提示缺少该类成员的默认构造函数。

I was trying to implement a move constructor for a class without a copy constructor. I got an error that the default constructor for a member of the class was missing.

这里有一个简单的例子来说明这一点：

Here's a trivial example to illustrate this:

struct A {
public:
        A() = delete;
        A(A const&) = delete;
        A(A &&a) {}
};

struct B {
        A a;
        B() = delete;
        B(B const&) = delete;
        B(B &&b) {}
};

尝试进行编译，我得到：

Trying to compile this, I get:

move_without_default.cc: In constructor ‘B::B(B&&)’:
move_without_default.cc:15:11: error: use of deleted function ‘A::A()’
  B(B &&b) {}
           ^
move_without_default.cc:6:2: note: declared here
  A() = delete;
  ^

为什么会出错？可以解决吗？

Why is this an error? Any way around it?

推荐答案

使用构造函数的初始化程序列表初始化 A 成员。按照编写程序的说法，正如编译器所说，move构造函数使用 A 的默认构造函数。

Use the constructor's initializer list to initialize the A member. As written, the move constructor uses, as the compiler says, the default constructor for A.

B(B&& b) : a(std::move(b.a)) {}

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