Raw strings starting/ending with backtick (original) (raw)

Jim Laskey james.laskey at oracle.com
Sat Nov 24 13:11:54 UTC 2018

There are several approaches but the simplest is using strip(). Example,

`` `abc` ``.strip()

Concat is another approach,

“`” + `abc` + “`”

Not perfect but other delimiter choices also have these edge cases.

Cheers,

— Jim

Sent from my iPad

On Nov 24, 2018, at 8:55 AM, Attila Kelemen <attila.kelemen85 at gmail.com> wrote:

Hi, Reading the JEP on raw string literals, I saw no mentions of the case when a string starts (or ends) with backtick. I guessed, that maybe the compiler will close the literal when it finds more than half the number of backticks than the beginning (nothing implied this behaviour just tried it and I know that it might be very suprising in other cases). I have tried with the latest early access compiler and (not too suprisingly) it didn't behave this way and simply failed when starting the literal with a backtick. My question is, of course: What is the strategy for this case? Or is it explicitly ignored as too much of an edge case (and left to the developer to deal with)? Thanks, Attila Kelemen

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