danmcb has asked for the wisdom of the Perl Monks concerning the following question:

Why can I do :

my @l = (1 .. 10);

and get the expected result, but not

my @l = (10 .. 1);

(not that it's really a problem as we have reverse(), but I just wondered.

Replies are listed 'Best First'.
Re: literal lists
by svenXY (Deacon) on Sep 06, 2005 at 09:30 UTC
    Well, the Camel Book justs says:
    "If the left value is greater than the right value, a null list is returned. (To produce a list in reverse order, see the reverse operator.)"
    No more explanation...
    Sorry, Sven
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      hah ... the answer is "cos they made it that way ..."

      fair enough.

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Re: literal lists
by phaylon (Curate) on Sep 06, 2005 at 11:30 UTC
    For instance: 
    my ( @a, @b );
@a = ( 1, 2, 3 );

foreach ( 0 .. $#a ) { }

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    Walks the values 0, 1 and 2. 
    foreach ( 0 .. $#b ) { }

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    does nothing. But $#b returns -1 because there are no values in it. If it would count down, It'd walk the values 0 and -1.
    Ordinary morality is for ordinary people. -- Aleister Crowley
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      true. In that case I wonder why $#b evaluates to -1, rather than undef?
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        I guess because undef in an integer context will be taken as '0' (albeit with a warning if you have asked for warnings), whereas -1 is not a valid array index: 

        @b = (a .. z);
                                                                      
+          
$#b = undef;
                                                                      
+          
print "@b";
                                                                      
+          
$#b = -1;
                                                                      
+          
print "@b";

        [download]

        /J\

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