regex - backwards?

[John]

Hi

$x =~s |a|b|;

That will replace the first 'a' by 'b' in $x.

Is there any obvious tweek so that it replaces the last occurrence of 'a'?

Regards
John

 

[Jürgen Exner]

$x =~s |a|b|;

That will replace the first 'a' by 'b' in $x.

Is there any obvious tweek so that it replaces the last occurrence of 'a'?

The most simple solutions are often the best solutions:
reverse;
s///;
reverse;
As long as you don't do that hundreds of thousands of times or you are
dealing with gigantic strings there shouldn't be much of a problem with
this trivial approach.

jue

 

[Danny Woods]

Hi

$x =~s |a|b|;

That will replace the first 'a' by 'b' in $x.

Is there any obvious tweek so that it replaces the last occurrence of 'a'?

Yes: regex lookahead:

$x =~ s|a(?=[^a]*$)|b|;

Can be read as "an 'a', followed by zero or more non-'a's up until the
end of line"

Cheers,
Danny.

 

[Jim Gibson]

Hi

$x =~s |a|b|;

That will replace the first 'a' by 'b' in $x.

Is there any obvious tweek so that it replaces the last occurrence of 'a'?

$x =~ s/(.*)a/$1b/;

should do it, taking advantage of the fact that the * modifier is
greedy.

 

[sln]

Hi

$x =~s |a|b|;

That will replace the first 'a' by 'b' in $x.

Is there any obvious tweek so that it replaces the last occurrence of 'a'?

Regards
John

If you don't want to use lookahead as Danny suggested,
which is probably easiest, a couple of alternatives:

$aaa =~ s/a ([^a]*) $ /b$1/x;
# or
substr ($aaa, pos($aaa)-1, 1) = 'b' if ($aaa =~ / .*a /xg);

Of these, lookahead may be quickest. Have to benchmark them.

-sln