C ++常量值引用 [英] C++ const lvalue references

问题描述

假设我有：

• 不可复制的A类


• B类作为成员，const A& a


• 一个函数 A GenerateA（）;

这是否意味着它应该是有效的：
B（GenerateA（））
？

即，const ref是否意味着generateA（）返回的A的副本没有被复制？这是否意味着只要B存在，返回的临时的范围就扩展了？

编辑：添加问题的意见：
是可接受的返回A&感谢！

< >解决方案

由于其他人已经说过， A GenerateA（）无法编译 A 不可复制。

关于const ref：no，临时的生存期不会延长到B的生存期。标准[12.2.5 ] states：

在构造函数的ctor-initializer（12.6.2）中临时绑定到引用成员，直到构造函数退出。 [...]函数返回语句（6.6.3）中返回值的临时绑定将一直存在，直到函数退出。

所以是的，一个临时的生命周期的延长存在于某些上下文中（有时真正有用：请参阅这篇文章），但不在您提供的一个。

最后一个问题，从 GenerateA（）（并将结果绑定到一个const引用不会有任何帮助）返回一个局部变量的引用是不合法的。 / p>

Assuming I have:

• class A which is non-copyable
• class B which has as a member, const A& a (and takes an A in its constructer and sets it in its initialization list)
• a function A GenerateA();

Does this mean that it should be valid to do: B(GenerateA()) ?

i.e, does the const ref mean that no copy of the A that generateA() returns is done? And does that mean that the scope of the returned temporary is extended for as long as B exists?

EDIT: Addon question from the comments: Is it acceptable to return a A& from GenerateA() to a local A, if the lvalue is a const A&?

Thanks!

解决方案

As it has already been stated by others, A GenerateA() cannot compile if A is not copyable.

Regarding the const ref : no, the lifetime of the temporary will not be extended to the lifetime of B. The standard [12.2.5] states :

A temporary bound to a reference member in a constructor's ctor-initializer (12.6.2) persists until the constructor exits. [...] A temporary bound to the returned value in a function return statement (6.6.3) persists until the function exits.

So yes, extension of the lifetime of a temporary exists in some contexts (and is sometime truly useful : see this article), but not in the one you presented.

Regarding your last question, it's not legal to return a reference to a local variable from GenerateA() (and binding the result to a const reference won't be of any help).

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