C++

Question

考虑这样的事情：

#include <iostream>

struct C {
    C(double x=0, double y=0): x(x) , y(y) {
        std::cout << "C ctor " << x << " " <<y << " "  << "\n";
    }
    double x, y;
};

struct B {
    B(double x=0, double y=0): x(x), y(y) {}
    double x, y;
};

struct A {
    B b[12];

    A() {
        b[2] = B(2.5, 14);
        b[4] = B(56.32,11.99);
    }
};


int main() {
    const B& b = A().b[4];
    C c(b.x, b.y);
}

当我用 -O0 编译时，我得到打印

C ctor 56.32 11.99

但是当我用 -O2 编译时我得到

 C ctor 0 0

我知道我们可以使用 const 引用来延长本地临时文件，所以像

const A& a = A();
const B& b = a.b;

完全合法。但我正在努力寻找为什么相同 mechanism/rule 不适用于任何类型的临时

的原因

编辑以供将来参考：

我使用的是 gcc 版本 6.3.0

Answer 1

A().b[4] 不是临时变量或 rvalue，这就是它不起作用的原因。虽然 A() 是临时的，但您正在创建对创建时存在的数组元素的引用。然后 dtor 触发 A()，这意味着稍后对 b.b 的访问变得有点不确定。您需要保留 A& 以确保 b 仍然有效。

const A& a = A();
const B& b = a.b[4];
C c(b.x, b.y);

Answer 2

您的代码应该格式正确，因为 temporaries

（强调我的）

Whenever a reference is bound to a temporary or to a subobject thereof, the lifetime of the temporary is extended to match the lifetime of the reference

给定A().b[4]，b[4]是b的子对象，数据成员b是temprorayA()的子对象，其生命周期应该被延长。

LIVE on clang10 with -O2
LIVE on gcc10 with -O2

顺便说一句：这似乎是 gcc 的 bug，它已被修复。

根据标准，[class.temporary]/6

The third context is when a reference is bound to a temporary object.³⁶ The temporary object to which the reference is bound or the temporary object that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference if the glvalue to which the reference is bound was obtained through one of the following:

...

[ Example:
template<typename T> using id = T;

int i = 1;
int&& a = id<int[3]>{1, 2, 3}[i];          // temporary array has same lifetime as a
const int& b = static_cast<const int&>(0); // temporary int has same lifetime as b
int&& c = cond ? id<int[3]>{1, 2, 3}[i] : static_cast<int&&>(0);
                                           // exactly one of the two temporaries is lifetime-extended
— end example ]

C++ - 使用 const 引用来延长临时成员，ok 还是 UB？

C++ - using const reference to prolong a member of a temporary, ok or UB?

language-lawyer

lifetime

temporary

reference-binding