On Nov 15, 6:52 pm, [EMAIL PROTECTED] (Kelly Jones) wrote:
> Consider:
>
> perl -le '$hash{"foo-bar"} = 1; print $hash{foo-bar}'
> [no result]
>
> perl -le '$hash{"foobar"} = 1; print $hash{foobar}'
> 1
>
> I sort of understand this: in the first script, Perl treats foo-bar as
> a subtraction, and
Dr.Ruud wrote:
> Rob Dixon schreef:
>> Kelly Jones:
>>>
>>> Consider:
>>>
>>> perl -le '$hash{"foo-bar"} = 1; print $hash{foo-bar}'
>>> [no result]
>>>
>>> perl -le '$hash{"foobar"} = 1; print $hash{foobar}'
>>> 1
>>>
>>> I sort of understand this: in the first script, Perl treats foo-bar
>>> as a
Rob Dixon schreef:
> Kelly Jones:
>> Consider:
>>
>> perl -le '$hash{"foo-bar"} = 1; print $hash{foo-bar}'
>> [no result]
>>
>> perl -le '$hash{"foobar"} = 1; print $hash{foobar}'
>> 1
>>
>> I sort of understand this: in the first script, Perl treats foo-bar
>> as a subtraction, and sets $hash{
On Sat, Nov 15, 2008 at 17:42, Jenda Krynicky <[EMAIL PROTECTED]> wrote:
snip
> The rule for automatic quoting within $hash{...} is "if it looks like
> word, it doesn't have to be quoted". And - is not in the list of word
> characters as far as Perl is concerned.
snip
Not exactly. A hyphen is all
Kelly Jones wrote:
> Consider:
>
> perl -le '$hash{"foo-bar"} = 1; print $hash{foo-bar}'
> [no result]
>
> perl -le '$hash{"foobar"} = 1; print $hash{foobar}'
> 1
>
> I sort of understand this: in the first script, Perl treats foo-bar as
> a subtraction, and sets $hash{0} to 1. In the second one
From: "Kelly Jones" <[EMAIL PROTECTED]>
> Consider:
>
> perl -le '$hash{"foo-bar"} = 1; print $hash{foo-bar}'
> [no result]
>
> perl -le '$hash{"foobar"} = 1; print $hash{foobar}'
> 1
>
> I sort of understand this: in the first script, Perl treats foo-bar as
> a subtraction, and sets $hash{0} to
On Sat, Nov 15, 2008 at 15:10, John W. Krahn <[EMAIL PROTECTED]> wrote:
snip
> s/pargmas/pragmas/;
snip
And I had paragams at one point. I need more (or better) sleep.
--
Chas. Owens
wonkden.net
The most important skill a programmer can have is the ability to read.
--
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Chas. Owens wrote:
On Sat, Nov 15, 2008 at 13:52, Kelly Jones <[EMAIL PROTECTED]> wrote:
Consider:
perl -le '$hash{"foo-bar"} = 1; print $hash{foo-bar}'
[no result]
perl -le '$hash{"foobar"} = 1; print $hash{foobar}'
1
I sort of understand this: in the first script, Perl treats foo-bar as
a s
Kelly Jones wrote:
Consider:
perl -le '$hash{"foo-bar"} = 1; print $hash{foo-bar}'
[no result]
Using warnings and/or strict may have helped:
$ perl -Mwarnings -le 'my %hash = ("foo-bar", 1); print $hash{foo-bar}'
Unquoted string "foo" may clash with future reserved word at -e line 1.
Argument
On Sat, Nov 15, 2008 at 13:52, Kelly Jones <[EMAIL PROTECTED]> wrote:
> Consider:
>
> perl -le '$hash{"foo-bar"} = 1; print $hash{foo-bar}'
> [no result]
>
> perl -le '$hash{"foobar"} = 1; print $hash{foobar}'
> 1
>
> I sort of understand this: in the first script, Perl treats foo-bar as
> a subtra
Consider:
perl -le '$hash{"foo-bar"} = 1; print $hash{foo-bar}'
[no result]
perl -le '$hash{"foobar"} = 1; print $hash{foobar}'
1
I sort of understand this: in the first script, Perl treats foo-bar as
a subtraction, and sets $hash{0} to 1. In the second one it assumes
you just left off some quot
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