template <typename T> class Obj { public: Obj<T>* doThis(); const Obj<T>* doThis() const {// return normal version of doThis()} }; In the example, I want the const version of doThis() to return normal version of doThis(). I don't think it's ok to just put return doThis() in definition of const version of doThis() because C++ may think this is recursion of const version of doThis().
Is there any way to explicitly tell C++ which function to call?
As Story Teller mentioned in the comment, don't do this, because this can lead to UB, e.g., in this simple case:
const Obj<int> obj; obj.doThis(); // UB since you will have to const_cast<Obj<int>&>(obj). The proper thing is to do it the other way:
template <typename T> class Obj { public: Obj<T>* doThis() { return const_cast<Obj<T>*>(std::as_const(*this).doThis()); } const Obj<T>* doThis() const { /* Actual implementation. */ } }; Note that this implies that the return value of doThis() const can be safely const_cast back to Obj<T>* (e.g., if you return this in doThis, this is safe).
If you cannot implement doThis() in the const-qualified version, then your code is probably flawed.
constmeans something here. It's a promise that the object will not be modified during the invocation of the function. If you call the non-const version you'll potentially break this promise. That can only cause problems.