触发器运算符/ readline-counter问题

Question

Reading this Perl: extract rows from 1 to n (Windows) I didn't understand the flip-flop-operator/readline-counter part.

perl -nE 'say $c if $c=1..3' my_file 

1
2
3E0

Could someone explain me more detailed where this output comes from?

Answer 1

To quote 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, even across calls to a subroutine that contains it. 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 until 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).

(emphasis added)