如何返回可以更改的会员？ [英] How to return member that can be changed?

问题描述

多线程环境。 Foo的内容可以是多线程。

Multi thread enviroment . The content of Foo can be multi thread.

class Foo
{
public:
   const A & getA() {return a_;} //has guard
   void setA(A newA){a_ = newA;} //has guard

private:
    A a_;
};

呼叫者：

A a  = foo.getA();

在另一个问题中，我问某人告诉我
如果您返回const&可以保证变量的生命周期会延长到引用的生命周期
，所以根据此，我不需要复制值，即使在调用后立即对foo调用setA也是安全的到getA，但是引起了很多反对，因此我觉得这是不正确的。

in another question that i asked someone told me that If you return const& it's guaranteed that life time of variable will be prolonged to lifetime of the reference , so according to this i dont need to copy the value and i safe even if call to setA on foo done right after i call to getA, but a lot of argument against it was araised , so i feel that this is not correct.

我想站在安全的一边，所以我更改了签名成为：

I want to be on the safe side so i change the signature to be :

A & getA() {return a_;}

，但是编译器仍然警告我，我已经引用了local变量。我不明白为什么，因为据我了解（cpp的新手），返回值是foo.a的副本，所以这是什么问题呢？

but the compiler still give me warning that i have reference to local variable. i dont understand why, because as far as i understand (new to cpp) the return value is a copy of foo.a, so what the problem with this?

i不担心a_内容的更改。（_ a.age = 4）。我担心调用set，并且调用方中的 a将是非法的

i am not worried about change of a_ content.(_a.age =4) . i worry about call to set and that my 'a' in the caller will be illegal

推荐答案

您需要更加小心你听。如果将临时对象立即绑定到const-reference，则唯一延长其寿命的时间。例如，像这样：

You need to be more careful who you listen to. The only time the lifetime of something gets extended if a temporary object is bound immediately to a const-reference. For example, like so:

Foo bar() { return Foo(); }

int main()
{
    Foo const & f = bar();

    /* stuff */

} // the object referred to by f is extended till here

您的情况并非如此。特别是返回const-reference 不会创建临时对象，因此这里没有任何东西可以延长寿命。特别是，以下内容绝对是 错误：

Your situation is nothing like that. In particular, returning a const-reference does not create a temporary object, so there's nothing here who's life gets prolonged. In particular, the following is definitely an error:

A const & bar() { Foo x; return x.getA(); }

int main()
{
    A const & a = bar(); // dangling reference; object dies upon return from Foo::getA()
}

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