Number notation

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

Replies are listed 'Best First'.
Re: Number notation by Sifmole (Chaplain) on Jun 20, 2001 at 22:15 UTC
Well.. `my %array = (1, 30, 2, 125);` [download] sets up a hash with 2 keys, 1 and 2. The values of which are 30 and 125 respectively. If you wrote it as such it might be more visible.. `my %array = ( 1 => 30, 2 => 125);` [download] You are querying the hash with keys of "02" and "2" from your splits. There is no key "02", since this is different from simply "2", thus "undef". There is a key "2" and you get the value of it "125". Your statement 2 == 02 is true, however the keys of a hash are treated as strings and "2" eq "02" is false.	[reply] [d/l] [select]
Re: Re: Number notation by Michalis (Pilgrim) on Jun 21, 2001 at 00:37 UTC
<Quote> Your statement 2 == 02 is true, however the keys of a hash are treated as strings and "2" eq "02" is false. </Quote> Actually that was the only thing I 've read so far, and I wasn't aware of, and that's the answer to the problem. Thanks. Well, learning while aging is a good thing anyway:)	[reply]
Re: Number notation by srawls (Friar) on Jun 20, 2001 at 22:15 UTC
because 2 == 02 is in numeric context. `$array{$var2}` is in string context. Just add int to the code like such: `print $array{int($var2)}` [download] The 15 year old, freshman programmer, Stephen Rawls	[reply] [d/l] [select]
Re: Re: Number notation by Michalis (Pilgrim) on Jun 21, 2001 at 00:41 UTC
Thanks, the usage of int() is a solution, and probably the one I'll follow (although, the hash is so small that i'ld be probably better to just add the extra keys :-)	[reply]
Re: Number notation by VSarkiss (Monsignor) on Jun 21, 2001 at 00:01 UTC
There are several good answer aboves. I just wanted to point out that you must have made a mistake transcribing the code. You're assigning values to `$string1` and `$string2`, but using `split` on a different variable `$string` (no number). The code as you have it here sets both `$var2` and `$var4` (and the others) to `undef`.	[reply] [d/l] [select]
Re: Re: Number notation by Michalis (Pilgrim) on Jun 21, 2001 at 00:33 UTC
Thanks. That was a typo, I meant $string1 and $string2 respectively	[reply]
Re: Number notation by cLive ;-) (Prior) on Jun 20, 2001 at 22:18 UTC
You have a hash with two name/value pairs. `$var1 = "2000"; $var2 = "02"; $var3 = "2000"; $var4 = "2"; $var2 does not get interpolated - ie, "02" is not the same as "2" in t +he context you used. To force numerical context: $array{$var2} = $array{"02"} = undefined $array{$var2+0} = $array{"2"} = 125` [download] 1 and 2 are the only KEYS in the hash, so others are undefined. And I would rename %array to %hash to avoid confusion with this. I suggest you read perldata document to familiarise yourself with Perl's data structures. cLive ;-) Update: - oops, misread (thanks Sifmole) - have amended above to explain.	[reply] [d/l]
Re: Re: Number notation by Sifmole (Chaplain) on Jun 20, 2001 at 22:21 UTC
cLive -- The exact confusion here is that "02" does NOT become "2" in $array{$var2}. If it did, then his result from $array{$var2} would be "125" not "undef".	[reply]


Syntactic Confectionery Delight
	PerlMonks