Does const on function parameters affect the signature and behavior in C++?

Hello everyone! I started learning C++ and encountered a confusing situation. When I write functions, I often use const for parameters to ensure they are not modified inside the function. For example: ```cpp void print(const std::string& str) { std::cout << str << std::endl; } ``` But recently I noticed that if I declare a function without const but with the same name and types, the compiler does not consider it an overload: ```cpp void print(const std::string& str) { /* ... */ } void print(std::string& str) { /* ... */ } // This is an overload void print(std::string str) { /* ... */ } // And this is an error or not? ``` Why is that? After all, const in a parameter is part of the type. But I heard that for parameters passed by value, const does not affect the signature. Is that true? And in general, is there any