How can I strip an arbitary number of leading spaces from a string?

[parsec]

How can I strip an arbitary number of leading spaces from a string?

Originally posted as a Categorized Question.

[japhy]

There is no need to use s/^\s*//. This will waste time on strings with no leading space. Use s/^\s+// at all times -- it obviously won't do anything if there's no whitespace to match.

A general rule of thumb is that s/X*// is slower than s/X+//, and in cases where the RHS (right-hand side) is something OTHER than "", chances are you really do mean X+.

[chromatic]

Use a regex with a quantifier. Use this one if you're not sure there will be any spaces at the start of the string:

$string =~ s/^\s*//;

Use this one if you know there will be at least one:

$string =~ s/^\s+//;

{QA Editors note: As japhy pointed out, the "+" quantifier is preferable over the "*" quantifier, even if it's unknown whether the string being tested has leading whitespace or not. With s/^\s+//;, strings lacking leading whitespace simply will be passed over. Furthermore, the "+" quantifier is faster than the "*" quantifier.}

[Anonymous Monk]

$string =~ s/^\s+//gm;

this will start at the beginning and only continue as long as there is whitespace left at the beginning of the string.

[Anonymous Monk]

HUGE thanks , dude :))

Originally posted as a Categorized Answer.