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. -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]