c++ - C++ 标准库必须支持对 friend 是谁挑剔的类吗？

Question

这个问题最容易用一个例子来说明，所以这里是:

像下面这样的代码是否保证有效，并且可以正确编译和运行？

(并不是所有的实现都能正确地编译它，但我想知道这是否是一个错误。)

#include <algorithm>
class Picky
{
    friend
        Picky *std::copy<Picky const *, Picky *>(Picky const *, Picky const *, Picky *);
    Picky &operator =(Picky const &) { return *this; }
public:
    Picky() { }
};

int main()
{
    Picky const a;
    Picky b;
    std::copy<Picky const *, Picky *>(&a, &a + 1, &b);
    return 0;
}

Answer 1

std::copy 需要一个输出迭代器 ([algorithms.general]/p5)；输出迭代器除其他外，要求 *r = o 有效([output.iterators]，表 108)——而不仅仅是“有时有效”或“在某些情况下有效”。

因为对于 Picky *p, a;，*p = a 在大多数情况下都是无效的，所以 Picky * 不是一个有效的输出迭代器。

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Hmm it'd be great if you could generalize your answer to other things beyond the particular example I gave. Like, for example, std::vector::push_back(T const &), or whatever.

与成员函数交 friend 是绝对不行的，因为你甚至不能保证有一个带有该签名的成员函数([member.functions]/p2，Stephan T. Lavavej 称之为 "STL Implementers Can Be Sneaky Rule" ):

An implementation may declare additional non-virtual member function signatures within a class:

• by adding arguments with default values to a member function signature¹⁸⁷ [Note: An implementation may not add arguments with default values to virtual, global, or non-member functions. — end note];
• by replacing a member function signature with default values by two or more member function signatures with equivalent behavior; and
• by adding a member function signature for a member function name.

_{¹⁸⁷ Hence, the address of a member function of a class in the C++ standard library has an unspecified type.}