regex question : ignoring first match in pattern search

[bcdixit]

I have the following piece of code,

#!/usr/bin/perl

$txt = "xxx yys";
$txt =~ s/\b/\!/g;
print "$txt\n";

basically I want to substitute at a word boundary a '!' character but not on the first word of the string or line.

the output that I am currently getting is

!xxx! !yys!

I want it to look something like this.

xxx! !yys!

Ideally I want it too be something like xxx!yys.

But I would highly appreciate if someone could give a hint of at least ignoring the first word boundary.

thanks
bcd

 

[ishnid]

This will not do the replacement if it's at the start of the string:

$txt =~ s/(?<!^)\b/\!/g;

 

[MacTommy]

If you want to just add exclamation marks at the end of each word, you could do:

$txt =~ s/(\w)\b/$1!/g;

This leaves you with an exlamation mark at the end of each line, but since this will always be the case, you might easily add a

$txt =~ s/!$//;

Maybe you can also do it in one go with some look-ahead assertion, but matching the end of the srting ($) in a look-ahead is a bit tricky I find...