Remember that pattern matches, by default, are *greedy*, so they match the longest string that conforms to the pattern. And unless you specify 'g' on the end, they only match the first substring that matches the pattern you specify.

So s/o*/e/ matches the first instance of "0 or more 'o's", and guess where that is? Right at the beginning of the string. When it says "match 0 or more", it means it =) So the result is "egood food".

With your second example, it only matches the first substring that's one or more 'o's, (the 'oo' in "good"), and replaces that with a single 'e' => 'ged food'.

HTH

perl -e 'print "How sweet does a rose smell? "; chomp ($n = <STDIN>); 
+$rose = "smells sweet to degree $n"; *other_name = *rose; print "$oth
+er_name\n"'
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There are a number of things going wrong here, not least of which is the fact that you are initialising $string1, $string2 and $string3 to an empty string - not what you think.

use strict would have caught this error.

But assuming that you do initialise all of your strings to "good food", the three results you get are "egood food", "ged food" and "geed feed" - which might also confuse you a bit. Let's look at the three regexes in a bit more detail.

s/o*/e/ says "look for zero or more os and convert them to one e". The first place you find zero or more os is at the start of the string, so that is changed to an e.

s/o+/e/ says "look for one or more os and convert them to one e". That finds the oo in "good" and converts them to an e.

s/o/e/g says "find one o and convert it to an e, then repeat that process for every o in the string".

"Perl makes the fun jobs fun
and the boring jobs bearable" - me

Note that if all you are trying to do is replace one character with another, it's much easier to use tr than s. For example:

$new_string = "good food";
$new_string =~ tr/od/ek/;
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$new_string = ( "good food" =~ tr/[od]/[ek]/; );
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No, it won't. tr/// in scalar context returns the number of characters translated; and the brackets are unnecessary. Not to mention, you're trying to modify a constant string. <code> ($new_string = "good food") =~ tr/od/ek/;

japhy -- Perl and Regex Hacker

$default_str = "good food";
$string1 = $default_string;

you need to use the same variable name ;) perl -w will help catch errors like that.

Update arturo of course, is right too, dunno if your problem was the misspelling or if that just happened when you put the code on perlmonks
                - Ant

$string = qq{
what I don't understand is why does pattern substitution
$string1 and $string2 does not gave me the end results
of "geed feed" ?
Doesn't 1 and 2 look for "o".. and for 1, it has a "*",
which means "match 0 or more", right?
and for 2, it has a "+", which means match 1 or more
times, right? thanks in advanced
};

$string =~ s/o/e/g;

print $string;
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#!/jpsama/bin/perl -w
$tks = `mount`;
$jpsama = $! if $!;
print $jpsama;