Lambda in parameters

[Cameron Simpson]

On 17Dec2020 23:52, Abdur-Rahmaan Janhangeer <arj.python at gmail.com> wrote:
>Here is a famous question's solution

These look like implementations of Lisp operators, which I've never 
found easy to remember. So I'll do this from first principles, but 
looking at the (uncommented) code.

This is some gratuitiously tricky code. Maybe that's needed for these 
operators. But there's a lot here, so let's unpick it:

>def cons(a, b):
>    def pair(f):
>        return f(a, b)
>    return pair

Cons returns "pair", a function which itself accepts a function "f", 
calls it with 2 values "a, b" and returns the result. For ultra 
trickiness, those 2 values "a, b" are the ones you passed to the initial 
call to cons().

This last bit is a closure: when you define a function, any nonlocal 
variables (those you use but never assign to) have the defining scope 
available for finding them. So the "pair" function gets "a" and "b" from 
those you passed to "cons".

Anyway, cons() returns a "pair" function hooked to the "a" and "b" you 
called it with.

>def car(c):
>    return c(lambda a, b: a)

The function accepts a function, and calls that function with "lambda a, 
b: a", which is itself a function which returns its first argument. You 
could write car like this:

    def car(c):
        def first(a, b):
            return a
        return c(first)

The lambda is just a way to write a simple single expression function.

>print(cons(1, 2)(lambda a, b: a))

What is "cons(1,2)". That returns a "pair" function hooked up to the a=1 
and b=2 values you supplied. And the pair function accepts a function of 
2 variables.

What does this do?

    cons(1, 2)(lambda a, b: a)

This takes the function returns by cons(1,2) and _calls_ that with a 
simple function which accepts 2 values and returns the first of them.

So:

    cons(1,2) => "pair function hooked to a=1 and b=2"

Then call:

    pair(lambda a, b: a)

which sets "f" to the lambda function, can calls that with (a,b). So it 
calls the lambda function with (1,2). Which returns 1.

>print(car(cons(1, 2)))

The "car" function pretty much just embodies the call-with-the-lambda.

>but i don't understand how lambda achieves this

If you rewite the lambda like this:

    def a_from_ab(a,b):
        return a

and then rewrite the first call to cons() like this:

    cons(1,2)(a_from_ab)

does it make any more sense?

Frankly, I think this is a terrible way to solve this problem, whatever 
the problem was supposed to be - that is not clear.

On the other hand, I presume it does implement the Lisp cons and car 
functions. I truly have no idea, I just remember these names from my 
brief brush with Lisp in the past.

Cheers,
Cameron Simpson <cs at cskk.id.au>