The expression `(1 .. 10 step 2)` returns an IntProgression, `('a' .. 'z' step 2)` returns a CharPro...

# stdlib

[natpryce]

The expression (1 .. 10 step 2) returns an IntProgression, ('a' .. 'z' step 2) returns a CharProgression, and there are equivalents for many built in types, but not for user-defined types.

Could the stdlib define the step function for ClosedRange<T>, to return a new stdlib type Progression<T> .?

[Sam]

Not every comparable can be added and subtracted, so it only really works for numeric types. What would step mean for e.g. a String? Unless it could take a lambda, like generateSequence...

[Sam]

infix fun ClosedRange<String>.step(step: (String) -> String): Iterable<String> =
    generateSequence(start, step).takeWhile { it in this }.asIterable()

for (s in "foo".."foooooo" step { it + "o" }) println(s)

I kind of like it! 😄

[Sam]

not entirely sure how sensible it is, though…

[natpryce]

I kind of like it too!

[natpryce]

But yeah, I get your point. There is no way to express “addable” in Kotlin.

[natpryce]

But a step function is a good compromise