C++ Function Overloading What is Function Overloading?
Example
Function overloading is having multiple functions declared in the same scope with the exact same name exist in the same place (known as scope) differing only in their signature, meaning the arguments they accept.
Suppose you are writing a series of functions for generalized printing capabilities, beginning with std::string:
void print(const std::string &str)
{
std::cout << "This is a string: " << str << std::endl;
}
This works fine, but suppose you want a function that also accepts an int and prints that too. You could write:
void print_int(int num)
{
std::cout << "This is an int: " << num << std::endl;
}
But because the two functions accept different parameters, you can simply write:
void print(int num)
{
std::cout << "This is an int: " << num << std::endl;
}
Now you have 2 functions, both named print, but with different signatures. One accepts std::string, the other one an int. Now you can call them without worrying about different names:
print("Hello world!"); //prints "This is a string: Hello world!"
print(1337); //prints "This is an int: 1337"
Instead of:
print("Hello world!");
print_int(1337);
When you have overloaded functions, the compiler infers which of the functions to call from the parameters you provide it. Care must be taken when writing function overloads. For example, with implicit type conversions:
void print(int num)
{
std::cout << "This is an int: " << num << std::endl;
}
void print(double num)
{
std::cout << "This is a double: " << num << std::endl;
}
Now it's not immediately clear which overload of print is called when you write:
print(5);
And you might need to give your compiler some clues, like:
print(static_cast<double>(5));
print(static_cast<int>(5));
print(5.0);
Some care also needs to be taken when writing overloads that accept optional parameters:
// WRONG CODE
void print(int num1, int num2 = 0) //num2 defaults to 0 if not included
{
std::cout << "These are ints: << num1 << " and " << num2 << std::endl;
}
void print(int num)
{
std::cout << "This is an int: " << num << std::endl;
}
Because there's no way for the compiler to tell if a call like print(17) is meant for the first or second function because of the optional second parameter, this will fail to compile.
