const引用数据成员，该成员绑定到构造函数中的该引用的临时初始化

考虑以下代码:

#include <iostream>

struct A {
    const char *name;

    A() : name("A") {
        std::cout << "A()\n";
    }

    virtual ~A() {
        std::cout << "~A()\n";
    }
};

class B {
    const A& a;
public:

    B() : a(A()) {
    };

    void print_data() const {
        std::cout << a.name << '\n';
    }

    ~B() {
        std::cout << "~B()\n";
    }
};

int main() {
    B b;
    b.print_data();
    return 0;
}

GCC 4.4的输出是:

The output from GCC 4.4 is:

A()〜A()A〜B()

A() ~A() A ~B()

这对我来说很奇怪.我本来希望要么将A的临时实例的副本绑定到B :: a，要么在〜B()期间销毁该临时实例.

This looks strange to me. I would have expected either a copy of the temporary instance of A being bound to the B::a or that temporary itself being destructed during ~B().

基本上，我认为B :: a在b的生命周期中始终是有效的引用.实际上，b.print_data()显然可以工作，并且编译器不会发出任何警告.

Basically, I thought B::a is always a valid reference during the life of b. In fact b.print_data() works apparently and the compiler does not give any warning.

标准的c ++ 98/03对这件事怎么说?

What does the standard c++98/03 say about this matter?