为什么不能映射:::指向const的指针？

Question

briefly, why does it not compile

#include <map>
int main()
{
    std::map<int*, char> m;
    const int *x = nullptr;
    m.find(x);
}

What could possibly be the reason for that not to be a valid piece of code?

Why does it matter to find whether it's a pointer or a pointer to a const??

Looks and smells like a bug...

And no thanks, no const_cast

Answer 1

Looks and smells like a bug...

Why would it be a bug?

The parameter of find is a const reference to the key type (const reference to int * ).

You can't bind const int * to such a reference, since const int * can't be implicitly converted to int * .

---

Since C++14 you could fix that by using a transparent comparator: std::map<int*, char, std::less<>> .

With a transparent comaprator find becomes a template. It will work with any parameter type, as long as it can be compared with the key type.

Answer 2

The default version of std::map<int*, char> uses less<int*> as comparator, which, in turn, has function ()(const int*, const int*) defined in it.

This function will not accept int*

However, if you use less<> , it will work:

std::map<int*, char, std::less<>> m;

The reason for that is that std::less<void> defines a template function,

template< class T, class U>
constexpr auto operator()( T&& lhs, U&& rhs ) const

And this works just fine when compares const vs non-const pointers.