Re: bitwise string operator question

String XOR does a bitwise XOR on a character by character bases. The bitwise representation of j is 01101010. The bitwise representation of a space is 00100000. XORing those gives you 01001010, the bitwise representation of J. And no, this is not a coincidence.

Abigail

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2Re: bitwise string operator question by jeffa (Bishop) on Aug 07, 2003 at 14:05 UTC
Thanks Abigail-II, i just learned something cool for the day: `perl -le'sub uc{@_[0]^" "};print uc($_)for(a..z)'` [download] Can't believe i hadn't picked that up before now ... DOH! Heh, my original one-liner did the bitwise XOR in a map ... i should know better than that. :P Let's try that again: `perl -le'sub flip{@_[0]^" "};print flip($_)for(a..z)'` [download] There, no ambiguity now. But ... the saddest part is that i learned this trick too late. My C++ Lab students last semester had to write a function that would flip case ... i really wish i could have blown their minds with this one. ;) jeffa L-LL-L--L-LL-L--L-LL-L-- -R--R-RR-R--R-RR-R--R-RR B--B--B--B--B--B--B--B-- H---H---H---H---H---H--- (the triplet paradiddle with high-hat)	[reply] [d/l] [select]
Re: bitwise string operator question by Abigail-II (Bishop) on Aug 07, 2003 at 15:05 UTC
That calls the buildin `uc`. You might want to call `::uc` instead. But beaware, it doesn't uppercase! It flips the case, uppercase to lowercase, and lowercase to uppercase. Abigail	[reply]
Re: 2Re: bitwise string operator question by Koschei (Monk) on Aug 08, 2003 at 07:15 UTC
Also be aware that, if you're using locales and such, uc() is aware how to transform accented characters and will DTRT.	[reply]