Is this the correct behaviour

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Kiran Kumar

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Apr 25, 2005, 8:12:17 AM4/25/05
to perl6-c...@perl.org
Hi ,
Is this behaviour correct when i print $_ ? ..


#!/usr/bin/pugs
use v6;


for (0 .. 8) -> $tmp
{
say "tmp is $tmp \n";
}

for (0 .. 8) -> $tmp
{
say " tmp is $tmp Spcl is $_ \n";
}

Output is

tmp is 0

tmp is 1

tmp is 2

tmp is 3

tmp is 4

tmp is 5

tmp is 6

tmp is 7

tmp is 8

tmp is 0 Spcl is 1

tmp is 2 Spcl is 3

tmp is 4 Spcl is 5

tmp is 6 Spcl is 7

tmp is 8 Spcl is


~Kiran

Luke Palmer

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Apr 25, 2005, 8:43:52 AM4/25/05
to Kiran Kumar, perl6-c...@perl.org
Kiran Kumar writes:
> Hi ,
> Is this behaviour correct when i print $_ ? ..
>
>
> #!/usr/bin/pugs
> use v6;
>
> for (0 .. 8) -> $tmp
> {
> say " tmp is $tmp Spcl is $_ \n";

> }
>
> tmp is 0 Spcl is 1
>
> tmp is 2 Spcl is 3
>
> tmp is 4 Spcl is 5
>
> tmp is 6 Spcl is 7
>
> tmp is 8 Spcl is

Nope. It seems that pugs is considering $tmp and the implicit usage of
$_ to specify two different parameters to the block. In fact, there's
only one: $tmp, and it is aliased to $_.

Luke

Aaron Sherman

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Apr 25, 2005, 4:43:39 PM4/25/05
to Luke Palmer, Perl6 Compiler List
On Mon, 2005-04-25 at 08:43, Luke Palmer wrote:

> Nope. It seems that pugs is considering $tmp and the implicit usage of
> $_ to specify two different parameters to the block. In fact, there's
> only one: $tmp, and it is aliased to $_.

Ah, and this was my problem too. Thanks, Luke or whoever fixed this one.

--
Aaron Sherman <a...@ajs.com>
Senior Systems Engineer and Toolsmith
"It's the sound of a satellite saying, 'get me down!'" -Shriekback


Kiran Kumar

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Apr 26, 2005, 4:16:45 AM4/26/05
to Aaron Sherman, Perl6 Compiler List
I Dont think this is fixed , Im still getting this Error .

for (0 .. 8) -> $tmp
{
say " tmp is $tmp Spcl is $_ \n";
}

Aaron Sherman

unread,
Apr 26, 2005, 6:51:20 AM4/26/05
to Kiran Kumar, Perl6 Compiler List
On Tue, 2005-04-26 at 13:46 +0530, Kiran Kumar wrote:
> I Dont think this is fixed , Im still getting this Error .
>
> for (0 .. 8) -> $tmp
> {
> say " tmp is $tmp Spcl is $_ \n";
> }

Interesting. Now, it's dependent on the use of $_. Take out that $_ and
it works fine... cool bug ;-)


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