这个C ++静态分析规则照原样有意义吗?

问题描述

我正在实现一些C ++静态分析规则，其中之一禁止某个函数返回对该函数的参考参数的引用或指针，即以下所有条件都不符合:

I'm implementing some C++ static analysis rules, and one of them prohibits a function from returning a reference or pointer to a reference parameter of the function, i.e. the following are all non-compliant:

int *f(int& x) { return &x; } // #1
const int *g(const int& x) { return &x; } // #2
int& h(int& x) { return x; } // #3
const int& m(const int& x) { return x; } // #4

给出的理由是无论引用参数是临时对象还是对该参数的引用，都是实现定义的行为."

The justification given for this is that "It is implementation-defined behaviour whether the reference parameter is a temporary object or a reference to the parameter."

但是，我对此感到困惑，因为C ++中的流运算符是以这种方式编写的，例如

I'm puzzled by this, however, because stream operators in C++ are written in this way, e.g.

std::ostream& operator<<(std::ostream& os, const X& x) {
    //...
    return os;
}

我认为我非常有信心C ++中的流运算符通常不会表现出实现定义的行为，那么到底是怎么回事?

I think I'm pretty confident that stream operators in C++ do not in general exhibit implementation-defined behaviour, so what's going on?

根据我目前的理解，我希望#1和#3定义明确，因为临时对象不能绑定到非const引用，因此int& x引用一个实物它的生存期超出了函数的范围，因此返回指向该对象的指针或引用就可以了.我希望#2会比较狡猾，因为可能会将const int& x绑定到一个临时目录，在这种情况下，尝试获取其地址似乎是一个错误的计划.我不确定#4-我的直觉是那也很容易躲闪，但我不确定.特别是，我不清楚在以下情况下会发生什么情况:

According to my understanding as it is at present, I would expect #1 and #3 to be well-defined, on the basis that temporaries cannot be bound to non-const references, so int& x refers to a real object that has lifetime beyond the scope of the function, hence returning a pointer or reference to that object is fine. I would expect #2 to be dodgy, because a temporary could have been bound to const int& x, in which case trying to take its address would seem a bad plan. I'm not sure about #4 - my gut feeling is that that's also potentially dodgy, but I'm not sure. In particular, I'm not clear on what would happen in the following case:

const int& m(const int& x) { return x; }
//...
const int& r = m(23);

推荐答案

您说过，#1和#3很好(尽管#1可以说是不好的风格).

As you say, #1 and #3 are fine (though #1 is arguably bad style).

由于#2相同的原因，#4令人迷惑.它允许传播对const的引用，超过其生存期的临时对象.

#4 is dodgy for the same reason #2 is; it allows propagating a const reference to a temporary past its lifetime.

让我们检查一下:

#include <iostream>

struct C {
  C() { std::cout << "C()\n"; }
  ~C() { std::cout << "~C()\n"; }
  C(const C &) { std::cout << "C(const C &)\n"; }
};

const C &foo(const C &c) { return c; }

int main() {
   const C &c = foo(C());
   std::cout << "c in scope\n";
}

这将输出:

C()
~C()
c in scope

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