C++ - 使用 const 引用来延长临时成员，好的还是 UB？

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);
}

当我使用 -O0 编译时，我得到打印

C ctor 56.32 11.99

但是当我用 -O2 编译时，我得到了

 C ctor 0 0

我知道我们可以使用 const 引用来延长本地临时时间，所以类似于

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

是完全合法的。但我正在努力寻找为什么相同的机制/规则不适用于任何类型的临时的原因

编辑以供将来引用:

我使用的是 gcc 版本 6.3.0

Answer 1

您的代码应该格式良好，因为对于 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 的子对象和数据成员 b 是 temproray A() 的子对象，其生命周期应该延长。

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

顺便说一句:这似乎是 gcc 的 bug已修复。

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 ]