来自方法的 C++ 引用变量

Question

这样做有意义吗:

class SomeClass
{
public:
    static Object getObject()
    {
        return Object("example")
    }
};

int main()
{
    const Object& myObject = SomeClass::getObject();

    // do something with myObject
}

或者这与只是获取对象一样

const Object myObject = SomeClass::getObject();

?

Answer 1

从 Herb Sutter 关于此主题的文章中，可以找到 here ，两种情况都是正确的:

Normally, a temporary object lasts only until the end of the full expression in which it appears. However, C++ deliberately specifies that binding a temporary object to a reference to const on the stack lengthens the lifetime of the temporary to the lifetime of the reference itself, and thus avoids what would otherwise be a common dangling-reference error.

但是如果你删除 const从情况1开始，就不行了。