英文:
Why does returning a reference to a local variable work in this case?
问题
以下是翻译的部分：
"我知道有很多类似的问题，但我找不到适合我的确切情况（操作符重载，通过复制传递）的答案。
据我所知，返回对局部变量的引用然后使用它应该导致未定义的行为。但在我的情况下似乎可以工作。这是否只是未定义行为有时也可以工作的情况，还是我忽视了某些事情？
// complex.cc
Complex& operator/(Complex lhs, const Complex& rhs)
{
return lhs /= rhs;
}
// 在main函数内部
complex::Complex a{3, -6};
complex::Complex b{1, -5};
complex::Complex res{a / b};
std::cout << a << '\n' << b << '\n' << res << std::endl;
我理解，当我将其传递给重载的运算符时，应该创建a的一个副本，然后返回对修改后副本的引用。我的理解是否错误，或者为什么在打印时可以工作？"
英文:
I know there are a ton of questions asking something similar, but I couldn't find anything for my exact situation (operator overload, pass by copy).
As far as I know, returning a reference to a local variable and then using it should result in undefined behaviour. But it seems to be working in my case. Is it simply a case of undefined behaviour can also mean it will work sometimes, or am I overlooking something?
// complex.cc
Complex& operator/(Complex lhs, const Complex& rhs)
{
return lhs /= rhs;
}
// inside main
complex::Complex a{3, -6};
complex::Complex b{1, -5};
complex::Complex res{a / b};
std::cout << a << '\n' << b << '\n' << res << std::endl;
As I understand it, shouldn't there be a copy of a created, when I pass it to the overloaded operator, which then returns a reference to the modified copy? Is that understanding wrong, or why does it work when I print it?
答案1
得分: 4
"但在我的情况下似乎正在工作。这只是一种未定义行为的情况，也可能意味着它有时会工作，还是我忽视了某些东西？"
"你没有。行为是未定义的，这意味着它可能看起来像是在工作。"
"正如我理解的那样，当我将它传递给重载运算符时，应该有一个已创建的副本，然后返回对修改后的副本的引用，对吗？"
"这也是正确的 - 并且当函数返回时，副本超出了范围，因此返回的引用是悬挂引用。"
英文:
> But it seems to be working in my case. Is it simply a case of undefined behaviour can also mean it will work sometimes, or am I overlooking something?
You are not. The behavior is undefined, which means that it may look like it's working.
> As I understand it, shouldn't there be a copy of a created, when I pass it to the overloaded operator, which then returns a reference to the modified copy?
That's also correct - and the copy goes out of scope when the function returns so the returned reference is a dangling reference.
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