How to avoid the copy when I return

 ̄綄美尐妖づ 提交于 2019-12-01 16:58:02

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

  • 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.

Check out return value optimization. A modern compiler will optimize this situation, and in straightforward situations like these, no copy will be made on any of the major compilers.

In principle, you could also create your object outside the function, and then call the function and pass the object to it by reference. That would be the old way of avoiding a copy, but it is unnecessary and undesirable now.

I usually work around this by having function signature as

void foo(set<int> *x)

Just pass it by reference or the other option is already mentioned in the comment.

Edit: I have changed the argument type to illustrate that x could be changed.

      set<int> s;
      foo(&s);

This is only preferred when you have an old compiler. I suppose that could be the case with some of the projects.

And, better thing to do will be Either to use move semantics with c++11. Or go ahead returning the container and look into RVO in modern compilers.

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!