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336

answers:

3

A lambda expression which takes a function (of one argument) and a number, and applies the function to twice the number.

+6  A: 

Applying the function to twice the number:

(lambda (f x) (f (* 2 x)))

Applying the function to the number twice (which is what you may have intended to ask):

(lambda (f x) (f (f x)))
Greg Hewgill
+3  A: 

Greg's answer is correct, but you might think about how you might break apart this problem to find the answer yourself. Here is one approach:

; A lambda expression
;(lambda () )

; which takes a function (of one argument) and a number
;(lambda (fun num) )

; and applies the function
;(lambda (fun num) (fun num))

; to twice the number
;(lambda (fun num) (fun (* 2 num)))

((lambda (fun num) (fun (* 2 num))) + 12)
grettke
+1  A: 

Here is another way to approach it:

Write a Contract, Purpose, and Header:

;; apply-double : function -> number -> any
;; to apply a given function to double a given number
(define (apply-double fun num) ...)

Write some Tests:

(= (apply-double identity 10) 20)
(= (apply-double - 15) -30)
(= (apply-double / 7) 1/14)

Define the function:

(define (apply-double fun num) 
  (fun (* 2 num)))

This is an abbreviation of the recipe here: http://www.htdp.org/2003-09-26/Book/

grettke