从函数返回对局部变量的常量引用 [英] Returning const reference to local variable from a function

问题描述

我有一些关于从函数返回对局部变量的引用的问题:

I have some questions on returning a reference to a local variable from a function:

class A {
public:
    A(int xx)
    : x(xx)
    {
        printf("A::A()
");
    }
};

const A& getA1()
{
    A a(5);
    return a;
}

A& getA2()
{
    A a(5);
    return a;
}

A getA3()
{
    A a(5);
    return a;
}

int main()
{
    const A& newA1 = getA1(); //1
    A& newA2 = getA2(); //2
    A& newA3 = getA3(); //3
}

我的问题是 =>

1. getA1() 的实现是否正确?我觉得这是不正确的，因为它返回的是局部变量或临时变量的地址.

1. Is the implementation of getA1() correct? I feel it is incorrect as it is returning the address of a local variable or temporary.

main (1,2,3) 中的哪些语句会导致未定义的行为?

Which of the statements in main (1,2,3) will lead to undefined behavior?

在const A&newA1 = getA1(); 标准是否保证在引用超出范围之前不会破坏由 const 引用的临时绑定?

In const A& newA1 = getA1(); does the standard guarantee that a temporary bound by a const reference will not be destroyed until the reference goes out of scope?

推荐答案

1.getA1() 实现是否正确?我觉得这是不正确的，因为它返回的是局部变量或临时变量的地址.

1. Is getA1() implementation correct ? I feel it is incorrect as it is returning address of local variable or temporary.

在您的程序中唯一正确的 getAx() 版本是 getA3().无论您以后如何使用它们，其他两个都有未定义的行为.

The only version of getAx() that is correct in your program is getA3(). Both of the others have undefined behaviour no matter how you use them later.

2.main (1,2,3) 中的哪个语句会导致未定义行为?

2. Which of the statements in main ( 1,2,3) will lead to undefined behavior ?

从某种意义上说，他们都没有.对于 1 和 2，未定义的行为是函数体的结果.对于最后一行，newA3 应该是一个编译错误，因为您不能将临时引用绑定到非 const 引用.

In one sense none of them. For 1 and 2 the undefined behaviour is as a result of the bodies of the functions. For the last line, newA3 should be a compile error as you cannot bind a temporary to a non const reference.

3.在 const A&newA1 = getA1(); 做标准保证由 const 临时绑定除非引用超出范围，否则引用不会被销毁?

3. In const A& newA1 = getA1(); does standard guarantees that temporary bound by a const reference will not be destroyed until the reference goes out of scope?

没有.下面是一个例子:

No. The following is an example of that:

A const & newConstA3 = getA3 ();

这里，getA3() 返回一个临时对象，该临时对象的生命周期现在绑定到对象 newConstA3.换句话说，临时文件将一直存在，直到 newConstA3 超出范围.

Here, getA3() returns a temporary and the lifetime of that temporary is now bound to the object newConstA3. In other words the temporary will exist until newConstA3 goes out of scope.

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