A lookbehind assertion, lookahead assertion, and well every other assertion
are zero width.
That means they do not advance the cursor.
So in the case of `` it asserts that the previous two
characters are `1:`
If that is a true assertion, then nothing happens.
If it is instead false, it causes the
Hi Sean, From the docs:
Lookbehind assertions:
https://docs.perl6.org/language/regexes#Lookbehind_assertions
"To check that a pattern appears after another pattern, use a
lookbehind assertion via the after assertion. This has the form:
"Therefore, to search for the string bar immediately pr
Perl 6 is doing the right thing. The dot matches any character. In
this case, matching the final ':'. The next bit of the regex says the
cursor has to be after 1:, and indeed, after matching the ':' the
cursor is after '1:', so the substitution succeeds.
Maybe you want s//x/ to append 'x' after '1
This seems like a bug, but I thought I'd ask here before reporting it.
$_ = '1:';
s/./x/;
.say;
This prints "1x". Is that what's supposed to happen somehow? I would have
thought that '1:' should only match a literal "1:", leaving nothing for the
dot to match.
