Simplest way of returning from a function without creating a copy

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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;
   };

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;
   }

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.

Posted 22-Mar-18 7:58am

David O'Neil

Updated 22-Mar-18 11:17am

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2 solutions

Solution 1

If you want dynamic memory allocation, yes unique_ptr can be the solution.
But even if you simply return function local structure variable from the function, this can be done without a copying. Caller function can pre-allocate stack space and callee will just fill this space. And this memory block will be used after the call without a copying.

Posted 22-Mar-18 10:28am

User 6901758

Comments

David O'Neil 22-Mar-18 16:32pm

Can you show an example, or point to a site that shows the technique in C++?

Thanks!

Or are you saying that it is where the caller sends the address to be filled in in the arguments? I am not looking for that type of solution.

Rick York 22-Mar-18 17:01pm

I believe what he is stating is you can pass a pointer into the function like this :

makeTrianglePointingDown( int x, int y, Triangle *pt );

and the pointer is the address of a stack object from the caller :

Triangle t;
makeTrianglePointingDown( x, y, &t );

David O'Neil 22-Mar-18 17:28pm

That's what I guessed, which I wasn't interested in. Have performed too many calls to 'GetClientRect' that way... Boring...

[no name] 23-Mar-18 3:22am

Hi! I meant the same stuff that in the solution #2. Compiler optimizes this through working with the same memory block in stack from the caller and callee.

David O'Neil 23-Mar-18 12:48pm

Thanks!

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CPallini · Accepted Answer · 2018-03-22T11:18:00

The simplest way is returning the object by value, the compiler optimizes that (NRVO).
Try

C++

#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;
}

Simplest way of returning from a function without creating a copy

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Solution 1

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Simplest way of returning from a function without creating a copy

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