如何避免复制时我回来 [英] How to avoid the copy when I return

问题描述

我有一个函数返回一个向量或集合：

I have a function which returns a vector or set:

set<int> foo() {
    set<int> bar;
    // create and massage bar
    return bar;
}

set<int> afoo = foo();

在这种情况下，我在函数foo（）中创建一个临时内存空间，然后
通过复制将其分配给 afoo 。我真的想避免这个副本，任何容易的方式我
可以做到这一点在C + + 11？我认为这与右值事件有关。

In this case, I create a temporary memory space in function foo(), and then assign it to afoo by copying. I really want to avoid this copy, any easy way I can do this in C++11? I think this has to do with the rvalue thing.

确定，更新到问题：如果我要返回一个由自己定义的对象，
不向量或集合的东西，这是否意味着我应该定义一个移动构造函数？
像这样：

OK, update to the question: If I am going to return an object defined by myself, not the vector or set thing, does that mean I should define a move constructor? like this:

class value_to_return {
  value_to_return (value_to_return && other) {
    // how to write it here? I think std::move is supposed to be used?
  }
}

THanks !!!

THanks!!!

推荐答案

Modem C ++编译器将实现：给定类型 T ：

Modem C++ compiler will implement: given a type T:

• 如果T有一个可访问的复制或移动构造函数，编译器可能会
  选择删除复制。这是所谓的（命名的）返回值
  优化（RVO），它甚至在C之前指定


• 否则，如果T有一个移动构造函数，T已移动（自C ++ 11 ）。


• 否则，如果T有复制构造函数，则会复制T。


• 否则会产生编译时错误。

• If T has an accessible copy or move constructor, the compiler may choose to elide the copy. This is the so-called (named) return value optimization (RVO), which was specified even before C++11 and is supported by most compilers.
• Otherwise, if T has a move constructor, T is moved(Since C++11).
• Otherwise, if T has a copy constructor, T is copied.
• Otherwise, a compile-time error is emitted.

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