Converting parameters when passing to functions (c++) - c++

Question

I am just starting to teach myself C++ and am having a hard time with function parameter passing. For example I am using my own print function where you simply put the string into the parameters and it logs it to the console.
//Print Function
void print(std::string message = "") {
std::cout << message << std::endl;
}
However because I declare it as a std::string variable if I pass it a number it will not print it. Ultimately I would like to make an input and print system like in Python. How to I go about this? Is there a way to convert the parameters to string? Or some other solution. Another function with similar problems is my input function:
//Input function (Enter does not work without text, space is not text)
std::string input(const char* message = "") {
std::cout << message;
std::string x;
std::cin >> x;
return x;
}
This does not allow the return to be an int witch makes calculations using the input harder. Any help is appreciated thanks in advance!
~ Moses

Answer 1

Besides template, if your compiler supports C++14, You can also use auto with lambda function. You can just write all these inside the main function.
auto print = [](const auto& message) {std::cout << message << std::endl;};
print(1); //1
print("AAA"); //AAA
Note that, unlike Python, when you want to print something, you don't need to convert it to a string first. As long as the thing you want to print has overloaded cout, You can simply cout it. And using template or auto doesn't change the fact that everything in C++ is statically typed, it's just that the compiler will create the different versions of overload functions for you automatically.
EDIT
As #Peter pointed out in the comment section, saying "cout is something that can be overloaded is flat-out wrong", and more accurate to say overloading the operator<< for the ostream and the corresponding class

Answer 2

Can++ templates are useful there.
//Print Function
template <typename T>
void print(const T& message) {
std::cout << message << std::endl;
}
void print() {
std::cout << std::endl;
}
Note, I removed a default argument value and used overloaded function. With passed empty argument type of template parameter can not be deduced. print does not work with containers and you need more efforts to print containers, in Python it works from box.
//Input function (Enter does not work without text, space is not text)
template <typename T>
T input(const char* message = "")
{
std::cout << message;
T x;
std::cin >> x;
return x;
}
Usage: int n = input<int>("Input number:");.

Answer 3

Alternatively I discovered a way to do this without using lambda:
void print(int message) {
std::cout << message << std::endl;
};
void print(float message) {
std::cout << message << std::endl;
};
void print(std::string message) {
std::cout << message << std::endl;
};
By making multiple functions with the same name it will use what ever one works, so any input (3.14, 8, "Hello") will all work and use corresponding function.