in reply to use y

Isn't y an alias to tr?

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Re^2: use y
by choroba (Cardinal) on Jul 14, 2016 at 23:00 UTC
    Tres, that's yue.

    ($q=q:Sq=~/;[c](.)(.)/;chr(-||-|5+lengthSq)`"S|oS2"`map{chr |+ord
}map{substrSq`S_+|`|}3E|-|`7**2-3:)=~y+S|`+$1,++print+eval$q,q,a,

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Re^2: use y
by Anonymous Monk on Jul 14, 2016 at 23:06 UTC
    No, the letter/string y is not an alias for letter/string tr, you're thinking of y/// operator which is an alias for tr/// operator.
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