John Benediktsson <mrj...@gmail.com>
writes:
> On Sun, Aug 21, 2011 at 12:49 PM, Arkady Rost <arkr...@gmail.com> wrote:
>> for i in rangeA {
>> for j in rangeB {
>> foo(param, i, j);
>> }
>> bar();
>> }
> This works for your particular task:
>
> { "1" "2" "3" } [
> { "a" "b" "c" } [ append ] with map " " join print
> ] each
Just in case you don't really need to do the "bar()" bit each loop, you
might also consider product-each from sequences.product. The first three
words below are equivalent to what John wrote (but have worse
names). The last word is slightly different, but might be much simpler
if you don't really need to run bar() *between* each iteration across
rangeB.
Rupert
USING: sequences sequences.product io ;
: with-second ( seq sec -- )
[ " " append append write ] curry each nl ;
: do-all ( seq1 seq2 -- )
[ with-second ] with each ;
: example-1 ( -- )
{ "a" "b" "c" } { "1" "2" "3" } do-all ;
: example-2 ( -- )
{ { "a" "b" "c" } { "1" "2" "3" } }
[ first2 append print ] product-each ;
Gives:
( scratchpad ) example-2
a1
b1
c1
a2
b2
c2
a3
b3
c3
pgpbidWwJDrDe.pgp
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