### Beginner question

Posted:

**Sat Aug 30, 2008 7:09 am**
by **cesar2000**

Hey I just started learning lisp, and there's a problem in my text-book I just can't seem to solve. Hope someone can help out.

We have a date in the form yyyymmdd. Define 3 functions year, month and day, which from a date will give the values year, month and day. e.g:

(year 19980220)

->1998

(month 19980220)

->2

(day 19980220)

->20

thanks.

### Re: Beginner question

Posted:

**Sat Aug 30, 2008 7:49 am**
by **cesar2000**

Ok I managed to figure out a way to solve it. Is there a better way?

(defun day (n) (mod n 100))

(defun month (n) (mod (/ (- n (day n)) 100) 100))

(defun year (n) (/ (- n (day n) (* (month n) 100)) 10000))

### Re: Beginner question

Posted:

**Sun Aug 31, 2008 9:14 pm**
by **Kohath**

I remember some comp.lang.lisp sage mentioning floor (

http://www.lispworks.com/documentation/HyperSpec/Body/f_floorc.htm). Thus directed, the definition of month could be:

- Code: Select all
`(defun month (n)`

(mod (floor n 100) 100))

Seeing that you now know about floor, the definition of year becomes even simpler than the above definition of month.

### Re: Beginner question

Posted:

**Fri Oct 10, 2008 8:56 am**
by **Jasper**

Are you sure they don't mean the string "yyyymmdd" then you can use

subseq.

- Code: Select all
`(defun year(str) (read-from-string(subseq str 0 4)))`

(defun month(str) (read-from-string(subseq str 4 6)))

(defun day(str)(read-from-string(subseq str 6 8)))

Of course, it assumes the string is long enough.

Hmm, is there an read-integer-from-string and such? (year "'hax0519") -> 'HAX ; 4