On Sat, Nov 03, 2001 at 05:49:37PM +0100, Thomas Hofer wrote: > Hi! > > As a perl beginner, I have a silly question about > array-slices-syntax: > > That's clear: > > perl -e '@a=(a,b,c);print "@a[1..2]\n"' > ==> b c > > perl -e '@a=(a,b,c);print "$a[1]\n"' > ==> b > > But why does $array[range] always give the first array-entry? > > perl -e '@a=(a,b,c);print "$a[1..2]\n"' > ==> a > > (OK, maybe it's useless, but I'd like to understand...)
Really? OK, then :-) You are using .. in a scalar context. This is totally different to using it in an array context. In scalar context it is the bistable operator which returns false while its first operand is false, then true until its second operand is true. Read more in perlop. When either operand to .. is a constant, which would otherwise not make sense, the operand is compared to $. - the number of the line currently being read. So, for you the .. operator is returning false, which is converted to 0. If you want to see what .. is returning, run this: $ echo "1\n2\n3\n4\n5\n6" | perl -ne 'print "<", scalar(3..5), ">\n"' <> <> <1> <2> <3E0> <> These return values are guaranteed. -- Paul Johnson - [EMAIL PROTECTED] http://www.pjcj.net -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
