On Thu, 10 Apr 2008 16:35:14 -0400, Jonathan Mast wrote:
> perl -ne 'print if 20 .. 50' file looks perfect but I totally don't
> understand it....
perldoc perlop:
In scalar context, ".." returns a boolean value. The operator is
bistable, like a flip-flop, and emulates the line-range (comma)
operator of sed, awk, and various editors. Each ".." operator
maintains its own boolean state. It is false as long as its left
operand is false. Once the left operand is true, the range operator
stays true until the right operand is true, AFTER which the range
operator becomes false again. It doesn't become false till the next
time the range operator is evaluated. It can test the right operand
and become false on the same evaluation it became true (as in awk), but
it still returns true once. If you don't want it to test the right
operand till the next evaluation, as in sed, just use three dots
("...") instead of two. In all other regards, "..." behaves just like
".." does.
The right operand is not evaluated while the operator is in the "false"
state, and the left operand is not evaluated while the operator is in
the "true" state. The precedence is a little lower than || and &&.
The value returned is either the empty string for false, or a sequence
number (beginning with 1) for true. The sequence number is reset for
each range encountered. The final sequence number in a range has the
string "E0" appended to it, which doesn't affect its numeric value, but
gives you something to search for if you want to exclude the endpoint.
You can exclude the beginning point by waiting for the sequence number
to be greater than 1.
If either operand of scalar ".." is a constant expression, that operand
is considered true if it is equal ("==") to the current input line
number (the $. variable).
--
Peter Scott
http://www.perlmedic.com/
http://www.perldebugged.com/
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