delete-second2 takes the same procedure from delete-second1, I just need to know how to use the if expression because you have to find out if the list has more than one item in order for it to delete anything. ________________________________________ From: danny....@gmail.com [danny....@gmail.com] on behalf of Danny Yoo [d...@hashcollision.org] Sent: Monday, September 17, 2012 12:44 PM To: Ashley Fowler Cc: users@racket-lang.org Subject: Re: [racket] Delete Second
On Mon, Sep 17, 2012 at 10:40 AM, Ashley Fowler <afowl...@broncos.uncfsu.edu> wrote: > Basically delete-second2 has to take a input of a list, if the list has more > than 1 item then it will delete the second item and return the new list, if > it doesnt have more than 1 item it will just return the original list. Let me be a bit more explicit. Is delete-second1 any different from delete-second2? * If it is the exact same problem, you get to just copy your answer from delete-second1 over to delete-second2. :) * If it is slightly different, then you can _reuse_ some of the hard work you did to solve delete-second1. * If there's no similarity whatsoever, then you have to design the function from scratch. Which one of these cases matches? ____________________ Racket Users list: http://lists.racket-lang.org/users
