Replacing template function foo(T*) by foo(T)

Question

In my code, I have a function:

template<typename T>
void foo (T*); // make sure that 'foo()' is passed only pointers

Now in new implementation, I am passing also a smart pointer (something like shared_ptr<>. So I have changed the signature of the function to,

template<typename T>
void foo (T);  // pointers or smart-pointers

The code shall work fine. However, is there any side effect I am missing ?

Answer 1

void foo(T) is pass-by-copy. T& (pass by reference) would be more efficient if sizeof(T) is non-small.

Answer 2

You could use enable_if to only allow pointers or instances of shared_ptr to compile if you want. That way you can get back the compile-time checking you had before (though, presumably, your functions implementation would do that as well).