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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