CODEWITHSUNDEEP
schemeracket
IntegrateThis
IntegrateThis

Reputation: 962

Scheme let bound statements

The following is in my class notes for Scheme:

(let ((x 2) (y 3))
  (let ((x 7) (z (+ x y)))
    (* z x)))

The answer yields 35. Can someone explain this to me? So on the 2nd line z(+x y) the x value seems to be 2 but after that (* z x) the x value is 7? Thanks a lot

Upvotes: 2

Views: 61

Answers (2)

Sylwester
Sylwester

Reputation: 48745

Perhaps the easier way to explain this is by looking at let as syntax sugar for anonymous procedure calls.

(let ((x 2) (y 3))
  (let ((x 7) (z (+ x y)))
    (* z x)))

Is the same as:

((lambda (x y)
   ((lambda (x z)
      (* z x))     ; first here is x 7
     7
     (+ x y)))     ; this x is from the outer
 2
 3)

Upvotes: 2

totoro
totoro

Reputation: 2456

(let ((x 2) (y 3))

Here the 1st let is still in charge until all values have been bound.

  (let ((x 7) (z (+ x y)))

Here the 2nd let is in charge.

    (* z x)))

If you want (x 7) to be used in (z (+ x y)) then try let*

TEST

(let ((x 2) (y 3))
  (let ((x 7) (z (+ x y)))
    (* z x)))

> 35

(let ((x 2) (y 3))
  (let* ((x 7) (z (+ x y)))
    (* z x)))

> 70

Hope it helps.

Upvotes: 4

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