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Perl: the Markov chain saw

Re^7: Perl list items

by robertw (Sexton)
on Jul 10, 2012 at 02:27 UTC ( #980807=note: print w/replies, xml ) Need Help??

in reply to Re^6: Perl list items
in thread Perl list items

Dave, thank you you are being very helpful, I am sorry I am quite new to perl and I don't understand it that well yet. Now when i do what you wrote:

my @five_only = @A[0..4]; print @five_only

I still get a list of about 300 numbers and letters does this mean I have to split since i need to make the string into independent items? I already tried that but it would not work this way unfortunately

my @values = split(',', @A); foreach my $val (@values) { d = d + 1); }

That would also be able to count however it seems that does not work, I tried the following just to get a new list of items that does function properly because when i requested te first five items i got a hundred numbers but that neither works since i got this response

<code> my @values = split(',', @A); foreach my $val (@values) { push (@vals, '$vals'); } <code/>


thanks to all you guys for your help, I am unfortunately quite slow in understanding and I have been wanting to write this program a particular time now, that is why i might appear quite impatient, my apologies for that

I am going to read the perlintro and the join split and others tomorrow

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Re^8: Perl list items
by davido (Archbishop) on Jul 10, 2012 at 07:03 UTC

    I'm sorry to say you've missed the point still. In your current code, @A contains only one element. That element contains a string that you created using join. You pass to join a list (or an array), and it returns a single scalar value that is essentially the concatenation of everything in the list you passed to it, glued together with the literal characters you specify as the first parameter to join.

    So, @A contains only one element, the way your original code was written. To put it another way:

    $A[0] eq "Some,value,with,commas,mixed,in"; $A[1] eq undef; # Nothing at all. $A[2] eq undef; # And same with $A[3], $A[4], and so on.

    $A[0] is the first element of @A, and the only element that contains any data. If you didn't want all of your data stored in the first element of @A, then you shouldn't have used join, because that's exactly what it did. It took a perfectly good list and turned it into a long string so that you could store it in $A[0].

    @values = split /,/, @A; would not work out either. That's because split doesn't want you to pass it a list as its second parameter (nor an array). If you do pass an array, it's evaluated in scalar context. You know what you get when you evaluate an array in scalar context? Nothing more than a count of how many elements are in the array. Know how many elements are in your array? One, because that's all you put there when you used join to populate the array (remember, join only returns one value, not a list).

    Back to split: What happens when you split "1" on the comma character? Nothing, no split occurs, because there are no commas in the value "1". What you get as a return value is a single element, "1". So now @values has $values[0] == 1, and nothing else.

    Moving on to the foreach: What happens when you iterate over a list that has a single element? You loop only one time. And on that one and only iteration, $val will be set to 1.

    That part about d = d + 1 ); doesn't compile; that's not Perl, so I can't help you with it. Furthermore, I have no idea how d would relate to the rest of your code. But it doesn't matter because the steps that led up to that loop were flawed to begin with.

    I can't do much more for you until you've taken the time to read perlintro, join, and split. After you've done that, I really think you will probably begin to get a sense of what is wrong in your code. And we'll be closer to speaking the same language with respect to what lists, contexts, arrays, scalars, and so on actually are.


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