>>>>> "JSJ" == J S John <phillyj...@gmail.com> writes:
JSJ> Hi all, I'm new to Perl. The only other language I know is
JSJ> Matlab/Octave and I'm still working my way around Linux. I am using
JSJ> Shlomi Fish's tutorial on perl until I get the llama book. I'm stuck
JSJ> on section [4.1. "+=" and friends ] [1]. I don't really understand the
JSJ> part "$a *= 2; $b += 1;" What is the *= mean?
others have given you the direct answer. here is the more general one
which you need to learn.
anytime you see an assignment operator which is always a two char op
ending in =, there is a simple rule to expand it.
$x += $y is $x = $x + $y
$x -= $y is $x = $x - $y
$x *= $y is $x = $x * $y
those (and other) assignment operators say to take the lvalue (the
location being assigned to) and make it the left side of the binary
operator in front of the =. take that result and assign it to the
lvalue.
i say lvalue (meaning something which can be on the left side of =,
i.e. it can be assigned to) because it can be expressions other than
scalar vars:
$totals{foo} += $foo_count ;
so += and friends are not only a shortcut when coding, they can be more
efficient since they only evaluate the lvalue one time and that can save
if that is hash/array or deeper access. and that keeps the logic cleaner
as well.
the most commonly used assignment ops are ||=, +=, .=, *= and maybe
-=.
||= is effectively a way to assign a default value. it doesn't work if
the lvalue already has either 0 or '' in it. the newer //= handles that
case as it only tests for defined. these are the same:
$arg //= 'default' ;
$arg = $arg // 'default' ;
and finally, when you don't know what a perl operator is, read perldoc
perlop. the section on assignment operators covers this and shows all
the legal types (16 of them).
uri
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