Yes, if $p were being assigned the return value of the flip-flop operator, as in the OP, then it would indeed be incremented by one on each iteration in which the flip-flop operator evaluates to true. But by using parentheses in:

if(($p=3)..5)
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you ensure that 3 is assigned to $p before the flip-flop operator is evaluated. Then the value of $p (now 3) is used in the evaluation of the flip-flop operator; and since 3 is “true” in Perl, the operator returns a sequence number (which is also true). That sequence number is not assigned to any variable, it’s just used to determine the outcome of the if statement.

Hope that helps,

Thanks Athanasius.

In your OP you already showed that it does.

it prints 3 for every iteration i.e.

#!/usr/bin/perl
##use warnings;
while(<>){
if(($p=3)..5)
{
print "p is $p\n";
}
}
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p is 3

p is 3

p is 3

p is 3

p is 3

p is 3

p is 3

p is 3

p is 3

p is 3

p is 3

p is 3

p is 3

p is 3

p is 3

p is 3

p is 3

p is 3

p is 3

In this code you are obviously always assigning 3.

What's the problem?

Cheers Rolf

_{(addicted to the Perl Programming Language and ☆☆☆☆ :)}