Re: literal lists

For instance:

my ( @a, @b );
@a = ( 1, 2, 3 );

foreach ( 0 .. $#a ) { }
[download]

Walks the values 0, 1 and 2.

foreach ( 0 .. $#b ) { }
[download]

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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Re^2: literal lists by danmcb (Monk) on Sep 06, 2005 at 12:03 UTC
true. In that case I wonder why $#b evaluates to -1, rather than undef?	[reply]
Re^3: literal lists by gellyfish (Monsignor) on Sep 06, 2005 at 12:10 UTC
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\	[reply] [d/l] [select]
Re^4: literal lists by danmcb (Monk) on Sep 06, 2005 at 12:42 UTC
That makes some sense. -1 is a valid index though, in normal use - it gives you the last element. But in the code you gave, the -1 indeed makes @b an empty array (I didn't know you could do that). Kind of like a quick way to truncate arrays.	[reply]