Function Composition Operator
c++, composition, std-function
Solution
Unfortunately, C++ does not allow operators to be overloaded for built-in types. Just as you cannot implement your own `operator*` for two `int`s, you cannot implement it for two function pointers, except if you represent these functions with the `std::function` type which is a class and therefore not a built-in type.
You can, however, instead of using an operator, simply use a function:
std::function<float(float)> dot(float(*func1)(float) , float(*func2)(float))
{
return [func1, func2](float x) { return func1(func2(x)); };
}
Problem
As a small exercise in functional programming in C++, I want to overload the `operator*` to allow me to compose 2 functions. What stumps me is that if I define my operator like this: ``` std::function<float(float)> operator*(std::function<float(float)> func1, std::function<float(float)> func2) { // ... } ``` I cannot use it with regular functions as in: ``` float f(float); float g(float); auto composite = f*g; ``` Because I get: `error: invalid operands of types ‘float(float)’ and ‘float(float)’ to binary ‘operator*’` If I try to add a function pointer only version as in: ``` std::function<float(float)> operator*(float(*func1)(float) , float(*func2)(float)) { return [func1, func2](float x) { return func1(func2(x)); }; } ``` I get this error: `error: ‘std::function<float(float)> operator*(float (*)(float), float (*)(float))’ must have an argument of class or enumerated type` I was able to get it to work by adding a `make_function` helper, but that really makes the code ugly. Is there a way to get this operator to automatically cast functions to `std::function` instances?