Need help understanding lambda (currying)

c#, currying, lambda

Solution

Let's first simplify the code:

Func<int, int> B(Func<int, int, int> f, int c)
{
    return x=>f(x, c);
}

This is just the same as:

class Locals
{
    public int c;
    public Func<int, int, int> f;
    public int Magic(int x) { return f(x, c); }
}
Func<int, int> B(Func<int, int, int> f, int c)
{
    Locals locals = new Locals();
    locals.f = f;
    locals.c = c;
    return locals.Magic;
}

Now is it clear what x refers to? x is the parameter to function "Magic".

Now you can use B like this:

Func<int, int, int> adder = (a, b)=>a+b;
Func<int, int> addTen = B(adder, 10);
int thirty = addTen(20);

Make sense? See what is happening here? We're taking a function of two parameters and "fixing" one of the parameters to a constant. So it becomes a function of one parameter.

The second example takes that one step further. Again, simplify to get rid of the cruft so that you can understand it more easily:

Func<int, Func<int, int>> B2(Func<int, int, int> f) 
{
    return y=>x=>f(x,y);
}

This is the same as

class Locals3
{
    public int y;
    public int Magic3(int x)
    {
       return x + this.y;
    }
}
class Locals2
{
    public Func<int, int, int> f;
    public Func<int, int> Magic2(int y)
    {
        Locals3 locals = new Locals3;
        locals.y = y;
        return locals.Magic3;
    }
}

Func<int, Func<int, int>> B2(Func<int, int, int> f) 
{
    Locals2 locals = new Locals2();
    locals.f = f;   
    return locals.Magic2;
}

So you say

Func<int, int, int> adder = (a, b)=>a+b;
Func<int, Func<int, int>> makeFixedAdder = B2(adder);
Func<int, int> add10 = makeFixedAdder(10);
int thirty = add10(20);

B is a parameter fixer. B2 makes a parameter fixer for you.

However, that's not the point of B2. The point of B2 is that:

adder(20, 10);

gives the same result as

B2(adder)(20)(10)

B2 turns one function of two parameters into two functions of one parameter each.

Make sense?

Problem

Original source