Avoiding const_cast when calling std::set<Type*>::find

Question

Is there any good way to obviate the const_cast below, while keeping const correctness?

Without const_cast the code below doesn't compile. set::find gets a const reference to the set's key type, so in our case it guarantees not to change the passed-in pointer value; however, nothing it guaranteed about not changing what the pointer points to.

class C {
public:
   std::set<int*> m_set;

   bool isPtrInSet(const int* ptr) const
   {
       return m_set.find(const_cast<int*>(ptr)) != m_set.end();
   }
};

Answer 1

Yes.

In C++14, you can use your own comparator that declares int const* as transparent. This would enable the template overload of find() that can compare keys against arbitrary types. See this related SO question. And here's Jonathan Wakely's explanation.

Answer 2

I want to explain the underlying logic of why this is impossible.

Suppose set<int*>::find(const int*) would be legitimate. Then you could do the following:

set<int*> s;
const int* p_const;
// fill s and p
auto it = s.find(p_const);
int* p = *it;

Hey presto! You transformed const int* to int* without performing const_cast.