Ah, now it's a different issue. It's a question of const member function versus non-const member function. The const qualifier will control which set of member functions you can use. Yes, in that case the const will make a difference, since one of the member functions return an int* while the other returns a const int*. Your original code (both in your latest and in the original post I criticized) let you manipulate an int* regardless.

But whether the class itself contains a pointer or not, and whether it was passed by value or by reference still is irrelevant to the issue (the pointer could be a global variable and it would still be the same). Passing by value, whether const or non-const doesn't allow you to directly modify the original object you passed to the function, since you're operating on a copy of the variable anyway.

One way or the other, if the variable contains a pointer (to non-const), you will be able to use that pointer to modify the pointee, whether the variable was passed by value or by reference, whether it is const or not. A copy of a pointer still points to the same address (shallow copy), not to a copy of the object at the original address (deep copy).

If you have struct Foo { int* ptr; }, then a const Foo object will contain a int* const, not a const int*. It's the pointer itself that is const, not the data it points to, so you can always manipulate the pointee regardless of the const-qualification of the object that contained the pointer. It's only if you want to re-target the pointer that you need to worry. In this case, if the object is const, you will obviously not be able to modify it.