关于c ++：比较typeid（）运算符中的两个type_info

Comparing two type_info from typeid() operator

比较两个typeid()结果可以吗？cppreference对此运算符有以下说明：

There is no guarantee that the same std::type_info instance will be
referred to by all evaluations of the typeid expression on the same
type, although std::type_info::hash_code of those type_info objects
would be identical, as would be their std::type_index.

1
2
3
4
5
6

const std::type_info& ti1 = typeid(A);
const std::type_info& ti2 = typeid(A);

assert(&ti1 == &ti2); // not guaranteed
assert(ti1.hash_code() == ti2.hash_code()); // guaranteed
assert(std::type_index(ti1) == std::type_index(ti2)); // guaranteed

我的理解是，返回是对类型为_info的静态l值的引用。这意味着&ti1==&ti2不能保证在相同类型中是相同的。而是说使用散列代码或std：：type_index类。但是，它没有提到是否直接比较类型：

1	ti1 == ti2;

保证是真实的。我以前就用过这个，文档是否暗示这是有保证的？