That was the full problem. It started with #2. I'm just confused on how to start #3. I know you have to use a if expression.
2. Write a procedure (delete-second1 LS) that takes a list LS of at least two items and returns the same list only with the second item deleted. You may assume the input list has at least two elements. Tests: (delete-second1 '(3 7)) ==> (3) (delete-second1 '(a b c d)) ==> (a c d) 3. Write a procedure (delete-second2 LS) that takes an arbitrary list LS of items and returns the same list only with the second item deleted if there is a second item, o.w. returns original input list. Hint: you will need to use a conditional. Tests: (delete-second2 '()) ==> () (delete-second2 '(3)) ==> (3) (delete-second2 '(3 7)) ==> (3) (delete-second2 '(a b c d)) ==> (a c d) ________________________________________ From: Matthias Felleisen [matth...@ccs.neu.edu] Sent: Monday, September 17, 2012 12:16 PM To: Ashley Fowler Cc: users@racket-lang.org Subject: Re: [racket] Delete Second On Sep 17, 2012, at 12:11 PM, Ashley Fowler wrote: > I just wanna know if I am starting the procedure out right? This is what I > got so far... > (define delete-second2(lambda(ls) > (if(>ls 1) Since you're asking a yes-no question, the answer has to be 'no'. Can you point us to the full problem statement please? The 'tests' specification looks wrong -- Matthias ____________________ Racket Users list: http://lists.racket-lang.org/users
