From Newsgroup: comp.lang.lisp

Jules Grosse wrote:

1) How can I reproduce the following behaviour:

(dotimes (x 10)
(dotimes (y 10)
(do-something-with x y)))

Using only one "do" construct?

Kenny Tilton wrote:

(do ((x 0 (if (= y 9)
(incf x)
x))
(y 0 (if (= y 9)
0 (incf y))))
((>= x 9))
(print (list x y)))

In "((>= x 9))", the >= should have been >.

Testing:

(dotimes (x 3)
(dotimes (y 3)
(print (list x y))))

(0 0)
(0 1)
(0 2)
(1 0)
(1 1)
(1 2)
(2 0)
(2 1)
(2 2)

(do ((x 0 (if (= y 2)
(incf x)
x))
(y 0 (if (= y 2)
0 (incf y))))
((>= x 2))
(print (list x y)))

(0 0)
(0 1)
(0 2)
(1 0)
(1 1)
(1 2)

The right way: [Tested with Gauche Scheme and SBCL.]

(do ((y 0 (if (= 2 y) 0 (+ 1 y)))
(x 0 (+ x (if (= 2 y) 1 0))))
((> x 2))
(print (list x y)))

(0 0)
(0 1)
(0 2)
(1 0)
(1 1)
(1 2)
(2 0)
(2 1)
(2 2)
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