C++ 入门第 5 版。 shared_ptr 和关联容器

Question

C++ 入门第 5 版。 Stanley lipmann 第 15 章 OOP 说：

std::multiset<std::shared_ptr<Quote>, decltype(compare)*> items{compare};
The elements in our multiset are shared_ptrs and there is no less-than operator for shared_ptr. As a result, we must provide our own comparison operation to order the elements (§11.2.2, p. 425). Here, we define a private static member, named compare, that compares the isbns of the objects to which the shared_ptrs point. We initialize our multiset to use this comparison function through an in-class initializer (§7.3.1, p. 274):"

但如果我尝试这样做：

// a class that doesn't define < operator
struct A
{
    int x = 0;
};


int main()
{
    std::shared_ptr<A> pa(make_shared<A>());
    std::shared_ptr<A> pb(make_shared<A>());

    cout << (pb < pa) << endl; // 0

}

尽管 class A 没有定义小于运算符，但为什么我的代码仍然有效？
问题是，在检查了关于 class std::shared_ptr 的 cppreference 之后，我发现它已经重载了关系运算符？！
我还针对 C++11 编译了代码并且仍然可以正常工作！
所以我希望有人给我解释一下书中的那一段。谢谢！

Answer 1

Why my code works although class A doesn't define less than operator?

因为当集合不包含 A.

类型的元素时，运算符 A 的内容无关紧要

该集合包含共享指针。

The thing is that after checking cppreference about class std::shared_ptr I've found out that it has overloaded relational operators?!

Cppreference 正确。

So I'd like someone to explain to me that paragraph in the book.

书错了。

C++ 入门第 5 版。 shared_ptr 和关联容器

C++ primer 5 ed. shared_ptr and associative containers

c++

shared-ptr