On Wednesday 16 January 2002 03:56 pm, rabs wrote:
> I am new to regualr expressions and becoming accqainted with the =~
> operator. It appears to me that the =~ allows me to match a pattern in a
> REGEX against a variable. As such it replaces the $_ varible.
>
> $name =~ /[rabs]/;
>
> mtaches with a string containing any of the following characters r a b
> s
>
> is this correct? I am quite confused....
The following text is from the perlop man-page (perldoc perlop) -- Steven
Binary "=~" binds a scalar expression to a pattern match.
Certain operations search or modify the string $_ by
default. This operator makes that kind of operation work
on some other string. The right argument is a search pat�
tern, substitution, or transliteration. The left argument
is what is supposed to be searched, substituted, or
transliterated instead of the default $_. When used in
scalar context, the return value generally indicates the
success of the operation. Behavior in list context
depends on the particular operator. See the Regexp Quote-
Like Operators entry elsewhere in this document for
details.
If the right argument is an expression rather than a
search pattern, substitution, or transliteration, it is
interpreted as a search pattern at run time. This can be
less efficient than an explicit search, because the pat�
tern must be compiled every time the expression is evalu�
ated.
Binary "!~" is just like "=~" except the return value is
negated in the logical sense.
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