将值或引用传递给需要存储副本的C ++构造函数？ [英] Pass by value or reference, to a C++ constructor that needs to store a copy?

问题描述

如果一个C ++（隐式或显式）值构造函数需要通过value或reference-to-const接受它的参数，当它需要以它的方式在它的对象中存储一个参数的副本吗？

Should a C++ (implicit or explicit) value constructor accept its parameter(s) by value or reference-to-const, when it needs to store a copy of the argument(s) in its object either way?

这是我能想到的最短的例子：

Here is the shortest example I can think of:

struct foo {
    bar _b;
    foo(bar [const&] b) // pass by value or reference-to-const?
        : _b(b) { }
};

这里的想法是，当foo对象被创建时，我想最小化对bar的拷贝构造函数的调用

The idea here is that I want to minimize the calls to bar's copy constructor when a foo object is created, in any of the various ways in which a foo object might get created.

请注意，我确实知道一些关于复制elision和（Named）的返回值优化，我已阅读想要速度？按值传递，但我不认为这篇文章直接针对这个用例。

Please note that I do know a little bit about copy elision and (Named) Return Value Optimization, and I have read "Want Speed? Pass by Value", however I don't think the article directly addresses this use case.

修改：我应该更具体。

假设我不知道 sizeof（bar），或者是否 bar 是一个基本的内置类型（ bar 可能是一个模板参数， foo 可以是类模板而不是类）。此外，不要假设 foo 的构造函数可以内联（或 bar '） 。假设我至少可能使用一个实现RVO的编译器。

Assume that I can't know the sizeof(bar), or whether or not bar is a fundamental, built-in type (bar may be a template parameter, and foo may be a class template instead of a class). Also, don't assume that foo's constructor can be inlined (or bar's, for that matter). Do assume that I at least might be using a compiler that implements RVO.

我想要的是有一个可能性（给编译器优化）这将调用 bar 的任何复制构造函数（即使在<$ c中执行 _b（b） $ c> foo 的初始化列表）：

What I would like is for there to be a possibility (given compiler optimizations) that a call like this will invoke no calls to bar's copy constructor whatsoever (even when executing _b(b) in foo's initialization list):

foo f = function_that_creates_and_returns_a_bar_object_using_rvo();

这是可能的（给定C ++ 98标准）因此，如果 foo 通过引用到const而不是通过值接受其参数，那么它是否可能工作？

Is there any possibility (given the C++98 standard) that this can be done, and if so, is it more or less likely to work if foo accepts its parameter by reference-to-const instead of by value?

推荐答案

在C ++ 98和C ++ 03中，应该传递 const& bar ，然后复制。在C ++ 0x，你应该传递 bar 然后做一个移动（提供 bar 有一个移动构造函数）。

In C++98 and C++03, you should pass const& bar and then copy. In C++0x, you should pass bar and then do a move (provided bar has a move constructor).

#include <utility>

struct foo
{
    bar _b;

    foo(bar b) : _b(std::move(b)) {}
};

如果使用lvalue参数构造foo，将调用复制构造函数以创建 b ，并将该副本移动到 _b 。如果你用一个右值参数构造foo， bar 的move构造函数将被调用来移动到 b 它将再次移动到 _b 。

If you construct foo with an lvalue parameter, the copy constructor will be called to create a copy b, and that copy will be moved into _b. If you construct foo with an rvalue parameter, bar's move constructor will be called to move into b, and then it will be moved again into _b.

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