Evaluate
(((lambda(x y) (lambda (x) (* x y))) 5 6) 10)in Scheme.
I am not sure how to do this actually!
((lambda (x y) (+ x x y)) 3 5)
Is fairly simple. Here x=3, y=5.
But in this the body is: (lambda (x) (* x y))) 5 6) and the parameter is 10?
So we evaluate separately? As in (lambda (x) (* x y))) 5 6) = (* 5 y) And then (((lambda(x y) (lambda (x) (* x y))) 5 6) 10) = (((lambda (x y) (* 5 y) 10))
But how can that be evaluated?
It is easier if you rename the parameter in the inner lambda:
(((lambda (x y) (lambda (z) (* z y))) 5 6) 10)
The outer lambda is applied to 5 and 6:
((lambda (x y) (lambda (z) (* z y))) 5 6)
This evaluates to
(lambda (z) (* z 6))
since yis bound to 6.
Note that x from the outer lambda (which is bound to 5) is never used.
This function is then applied
((lambda (z) (* z 6)) 10)
which yields 60.
(((lambda (x y) (lambda (x) (* x y))) 5 6) 10)
is
( ( (lambda (x y)
(lambda (x) (* x y))
)
5
6)
10)
is
See? The inner lambda is returned as a value, as result of the application of the outer lambda to arguments 5 and 6. The new value is enclosed in its defining environment, where x=5 and y=6 -- such lambda is otherwise known as closure:
It is then applied to 10:
(((lambda (x y) (lambda (x) (* x y))) 5 6) 10)
= ( (let ((x 5)
(y 6))
(lambda (x) (* x y)))
10)
= (let ((x 5)
(y 6))
( (lambda (x) (* x y))
10))
= (let ((x 5)
(y 6))
(let ((x 10))
(* x y)))
= (* 10 6)
= 60
Let's start with (lambda(x y) (lambda (x) (* x y))). This is a function expression that takes two parameters and returns another function expression. For convenience, I will refer to the entire expression as f.
Now consider (f 5 6). This returns another lambda, (lambda (x) (* x y), with y bound to the value passed in. However, the inner x parameter shadows the outer one, so the result is (f 5 6) = (lambda (x) (* x 6).
From here, we can directly evaluate ((lambda (x) (* x 6)) 10). This yields a final result of 60.
