Functional programming in C++. Implementing f(a)(b)(c)
c++, c++11, currying, functional-programming, std-function
Solution
You can do it by having your function `f` return a functor, i.e., an object that implements `operator()`. Here is one way to do it:
struct sum
{
double val;
sum(double a) : val(a) {}
sum operator()(double a) { return val + a; }
operator double() const { return val; }
};
sum f(double a)
{
return a;
}
Example
Link
int main()
{
std::cout << f(1)(2)(3)(4) << std::endl;
}
Template version
You can even write a templated version that will let the compiler deduce the type. Try it here.
template <class T>
struct sum
{
T val;
sum(T a) : val(a) {}
template <class T2>
auto operator()(T2 a) -> sum<decltype(val + a)> { return val + a; }
operator T() const { return val; }
};
template <class T>
sum<T> f(T a)
{
return a;
}
Example
In this example, `T` will ultimately resolve to `double`:
std::cout << f(1)(2.5)(3.1f)(4) << std::endl;
Problem
I have been getting into the basics of functional programming with C++. I am trying to make a function `f(a)(b)(c)` that will return `a + b + c`. I successfully implemented the function `f(a)(b)` which returns a + b. Here is the code for it: ``` std::function<double(double)> plus2(double a){ return[a](double b){return a + b; }; } ``` I just cannot figure out how to implement the function `f(a)(b)(c)` which as I previously stated should return `a + b + c`.