Need help understanding lambda (currying)

Question

i am reading Accelerated C# i don't really understand the following code:

public static Func<TArg1, TResult> Bind2nd<TArg1, TArg2, TResult> (
    this Func<TArg1, TArg2, TResult> func,
    TArg2 constant ) 
{
    return (x) => func( x, constant );
}

in the last line what is x referring to? and there's another:

public static Func<TArg2, Func<TArg1, TResult>> Bind2nd<TArg1, TArg2, TResult>
( this Func<TArg1, TArg2, TResult> func )
{
    return (y) => (x) => func( x, y );
}

How do i evaluate this? (y) => (x) => func( x, y ) what is passed where ... it does confusing.

Answer 1

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?

Answer 2

x is the parameter of the lambda, it is of type TArg1.

It can be helpful to pronounce the => as "maps to" as in "x maps to a new function with a constant of type TArg2 substituted into the original function delegate, func."

Answer 3

The variable x is an unbound variable. That represents an argument to the returned function from calling Bind2nd.

A few hours with Scheme would help you here, but try this.

When you call Bind2nd the returned result is a function. That function is defined as

(x) => func (x, constant)

Now that you have the above assigned to a variable, lets say lambda, you can call that function via the lambda variable

lambda(x);

The x defined in Bind2nd is just a variable that represents an argument to the function that will be returned to you.