如果返回参数，可以复制省略号吗？ [英] Copy elision possible if returning parameter?

问题描述

考虑一个函数，该函数按值获取对象，对该对象执行一些操作并返回该对象，例如：

Consider a function that takes an object by value, performs some operations on it and return that object, for example:

std::string MyToUpper (std::string s)
{
      std::transform(s.begin(), s.end(), s.begin(), std::toupper);
      return s;
}

现在您可以临时调用此函数：

Now you call this function with a temporary:

std::string Myupperstring = MyToUpper("text");

从概念上讲，不需要复制。在这种情况下，现代编译器是否可以删除所有副本？如果没有，只有动作吗？那么这种情况呢？

Conceptually there is no copy needed. Are modern Compilers able to elide all copies in this case? If not, are there only moves? What about this case:

std::string Mylowerstring("text");
std::string Myupperstring = MyToUpper(std::move(Mylowerstring));

推荐答案

最多可以删除两个概念副本之一。如果将临时函数传递给该函数，则根据C ++ 11 12.8 / 31的第三个项目符号，可以删除该副本：

At most one of the two conceptual copies can be elided. If you pass a temporary to the function, then that copy can be elided, per the third bullet of C++11 12.8/31:

当要复制/移动临时类对象...时，可以省略复制/移动操作

when a temporary class object ... would be copied/moved ..., the copy/move operation can be omitted

回报不容忽视；只能针对临时变量（根据上面引用的规则）或局部变量，针对第一个项目符号：

The return can't be elided; that can only be done for temporaries (per the rule quoted above) or local variables, per the first bullet:

在return语句中。 ..当表达式是
非易失性自动对象的名称（函数或捕获子句参数以外）...时，可以省略复制/移动操作

in a return statement ... when the expression is the name of a non-volatile automatic object (other than a function or catch-clause parameter) ... the copy/move operation can be omitted

在没有省略的情况下，将返回值视为 rvalues 并在可能的情况下将其移动（并且可能在此处）;如果函数参数是 rvalues ，则会移动它们，就像在您的两个示例中一样。

In the absence of elision, return values are treated as rvalues and moved if possible (and it is possible here); function arguments are moved if they are rvalues, as they are in both your examples.

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