RosettaCodeData/Task/Substring/Aikido/substring.aikido

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2013-04-11 01:07:29 -07:00
const str = "abcdefg"
var n = 2
var m = 3
println (str[n:n+m-1]) // pos 2 length 3
println (str[n:]) // pos 2 to end
println (str >> 1) // remove last character
var p = find (str, 'c')
println (str[p:p+m-1]) // from pos of p length 3
var s = find (str, "bc")
println (str[s, s+m-1]) // pos of bc length 3