RVO何时启动？ [英] When should RVO kick-in?

问题描述

从以下代码，如果RVO发生，我希望看到2个地址指向同一位置，但是不是这样（我的编译器是MS VC9.0）

From the following code, If RVO has happened, I expect to see the 2 addresses pointing to the same location, however this is not the case (my compiler is MS VC9.0)

#include <iostream>
#include <string>

std::string foo(std::string& s)
{
   std::cout << "address: " << (unsigned int)(&s) << std::endl;
   return s;
}

int main()
{
   std::string base = "abc";
   const std::string& s = foo(base);
   std::cout << "address: " << (unsigned int)(&s) << std::endl;
   std::cout << s << std::endl;
   return 0;
}

RVO在什么情况下会发生？

Under what conditions should RVO be happening?

btw，我的问题是关于以下讨论：http://cpp-next.com/archive/2009/08/want-speed-pass-by-value/

btw, I'm basing my question on the following discussion: http://cpp-next.com/archive/2009/08/want-speed-pass-by-value/

推荐答案

RVO通常适用于您返回未命名的临时数据，但如果您返回之前创建的对象，则不会。

RVO generally applies when you return an unnamed temporary, but not if you return a previously created object.

std::string foo() {
  return std::string("hello world"); // RVO
}

std::string foo() {
  std::string str("hello world");
  bar();
  return str; // Not RVO
}

std::string foo(std::string str) {
  return str; // Not RVO
}

更普遍的版本是NRVO ，这也适用于命名变量。

A more general version is NRVO (Named return value optimization), which also works on named variables.

std::string foo() {
  std::string str("hello world");
  bar();
  return str; // NRVO
}

std::string foo(std::string str) {
  return str; // Not NRVO, as far as I know. The string is constructed outside the function itself, and that construction may be elided by the compiler for other reasons.
}

std::string foo(std::string str) {
  std::string ret;
  swap(ret, str);
  return ret; // NRVO. We're returning the named variable created in the function
}

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