free variables in procedures

eliminating a parameter

• helper’s parameter x is bound to the same value on every call
• so can eliminate the parameter
• now x in the body of helper refers to the parameter of exp
  (define (exp x k)
  
  (define (helper k s)(if (= k 0) s
  (helper x (- k 1) (* x s))))
  (helper k 1))

note

• x is now a free variable of the procedure helper
• so helper would not work elsewhere if we cut and pasted the code

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