perlop also describes the behavior of the range operator in list context, which is simply to return a list of values counting by one from the left operand to the right operand.

'z' .. 'a' is a special case. When the range operator is using magical autoincrement (like $var++), it stops when the string is greater than the right operand, or when the string is longer than the right operand. Thus, you'd get the same result from 'z' .. 'z' or 'z' .. '.'.

Okay, now I see the difference between list and scalar context...I thought the whole section on the range operator referred to it in list context(and I tried to make sure that I carefully read the docs...DOH!). It is definitely more clear now.

Still a bit confused though. After re-reading, I tried a few more examples. The goal of each is to print the numbers 5,6,7,8,9,10, one to a line.

@arr = (1 .. 20);
foreach $i ( 0 .. $#arr ) {
    if ( $i == (5 .. 10) ) {
        print "$i\n";
    }
}
# output was "0\n"

@arr = (1 .. 20);
foreach $i ( 0 .. $#arr ) {
    if ( $i == 5 .. 10 ) {
        print "$i\n";
    }
}
# output was 5-19, one integer to a line

@arr = (1 .. 20);
foreach $i ( 0 .. $#arr ) {
    if ( $i = (5 .. 10) ) {
        print "$i\n";
    }
}
# no output - '$i = (5 .. 10)' never evaluated true
[download]

Maybe I am misunderstanding how to use the operator is scalar context. The examples in the docs don't really help. Of all the examples listed, I thought the first was going to work. I'm really not understanding why it didn't.

Thanks for your help to this point.

Amel - f.k.a. - kel

In scalar context, the range operator yields true or false. So comparing $i == (5..10) does not make sense. The code you want is:

  for $i (@arr) {
    if ($i==5 .. $i==10) {
      print "$i\n";
    }
  }
[download]

Or:

  for $i (@arr) {
    print "$i\n" if $i==5 .. $i==10;
  }
[download]

--
Mark Dominus
Perl Paraphernalia

Okay, your usage of the range operator in a scalar context isn't quite correct.

I think what you're looking for is this:

foreach $i ( 0 .. 19 ) {
  if ($i == 5 .. $i == 10) {
    print "$i\n";
  }
}
[download]

You need to put the full comparison on each side of the range operator. Otherwise, you're doing something that's the equivalent of if ($string eq 'foo' || 'bar' || 'baz') {.

There's one extra bit of info from perlop:

If either operand of scalar ".." is a constant expression, that operand is implicitly compared to the $. variable, the current line number.

So your snippets are doing implicit comparisons against $.:

$i == (5 .. 10) compares $i to the result of the flip-flop, which flips when $. equals 5 and flops when $. equals 10.
$i == 5 .. 10 flips when $i equals 5 and flops when $. equals 10.
$i = (5 .. 10) assigns to $i the result of the flip-flop, which flips $. equals 5 and flops when $. equals 10.</code>

In a scalar context it's a flip flop operator as you described.

And for() gives a list context.

You might want to pop back to perlop and re-read how it acts differently in different contexts.

In my perlop documentation it states: 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. This is the opposite of what you wrote. Perhaps you misread it.