Function that takes one explicit type

Question

How can I make a function that accepts, for example, a size_t but does NOT accept an int or any other implicitly convertible type?

Answer 1

Use overload:

template <typename T> void foo(T) = delete;

void foo(std::size_t t) {
    // ...    
}

Answer 2

#include <type_traits>
#include <cstddef>

template <typename..., typename T, std::enable_if_t<std::is_same_v<T, std::size_t>>*...>
void f(T x) {}