在不创建副本的情况下从函数返回的最简单方法

问题描述

我刚刚进入移动语义，从我的阅读中，它只适用于一个类' new 的内容。我是否遗漏了某些东西，或者这样做是否如此，如果我想在从函数返回时消除元素的复制，一种方法是使用 std :: unique_ptr ？例如，给定 struct ：

I'm just getting into move semantics, and from my reading it is only useful when a class 'new's something within it. Am I missing something, or does that make it so that if I want to eliminate the copying of an element upon returning from a function, one way to do so is to use a std::unique_ptr? For example, given the struct:

struct Triangle {
   POINT p1;
   POINT p2;
   POINT p3;
   };

如果我想从创建一个函数的函数返回，最好的方法就是如下：

If I want to return from a function that creates one, the best way to do so is something like the following:

std:unique_ptr<MouseHelper::Triangle> MouseHelper::makeTrianglePointingDown(int x, int y) {
   //The following is hard coded for the sizes I'm using in the example
   const int halfSize = 5;
   std:unique_ptr<Triangle> t(new Triangle);
   t->p1.x = x-halfSize;
   t->p1.y = y-halfSize-halfSize;

   t->p2.x = x+halfSize;
   t->p2.y = y-halfSize-halfSize;

   t->p3.x = x;
   t->p3.y = y-2;

   return t;
   }

否则，我必须将' Triangle '变成'一个类， new 该类中的结构，然后创建一个合适的移动构造函数？



谢谢！



我的尝试：



阅读。上面的代码可以工作。

Otherwise, I must make 'Triangle' into a class, and new a structure within that class, and then create an appropriate move constructor?

Thanks!

What I have tried:

Reading. And the above code works.

推荐答案

#include <iostream>
using namespace std;

struct Point
{
  int x, y;
};

struct Triangle
{
  Point p1,p2,p3;
  Triangle(){cout << "Triangle ctor" << endl;}
};

Triangle makeTrianglePointingDown(int x, int y)
{
   const int halfSize = 5;

   Triangle t;
   t.p1.x = x-halfSize;
   t.p1.y = y-halfSize-halfSize;

   t.p2.x = x+halfSize;
   t.p2.y = y-halfSize-halfSize;

   t.p3.x = x;
   t.p3.y = y-2;

   return t;
}

int main()
{
  Triangle tpd  = makeTrianglePointingDown(10,10);
  cout << tpd.p1.x << " " << tpd.p1.y << " ";
  cout << tpd.p2.x << " " << tpd.p2.y << " ";
  cout << tpd.p3.x << " " << tpd.p3.y << endl;
}

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