This works for your particular task:
{ "1" "2" "3" } [
{ "a" "b" "c" } [ append ] with map " " join print
] each
On Sun, Aug 21, 2011 at 12:49 PM, Arkady Rost <[email protected]> wrote:
> Hi! I'm looking for a more elegant solution of the following task.
>
> for i in rangeA {
> for j in rangeB {
> foo(param, i, j);
> }
> bar();
> }
>
> I need to translate this to factor.
>
> foo(param, y, x) <=> x y param foo
>
> I have a solution:
> rangeA rangeB param [ foo ] curry [ swapd [ call ] 2curry each bar ] 2curry
> each
>
> For example,
> rangeA = { "a" "b" "c" }
> rangeB = { "1" "2" "3" }
> param = " "
> foo = append append write ;
> bar = "" print ;
>
> { "a" "b" "c" } { "1" "2" "3" } " " [ append append write ] curry [ swapd
> [ call ] 2curry each "" print ] 2curry each
>
> Result is :
> 1a 2a 3a
> 1b 2b 3b
> 1c 2c 3c
>
> I think my solution is bloated with partial applications. Maybe someone can
> provide a better way to solve this task.
>
> Cheers,
> Ark. Rost
>
>
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