Hi,
i have implemented 50 shell program to test. now i want to run all these
shell program one by one(in short test automation) (once first program
getting executed then next will start). please let me know how to do this?
in UNIX any utility or to run the same.
any help will be really appreciat
Hans Ginzel wrote:
Hello!
Hello,
Is there a shorter way to write $a = ! $a, please?
No.
John
--
Any intelligent fool can make things bigger and
more complex... It takes a touch of genius -
and a lot of courage to move in the opposite
direction. -- Albert Einstein
--
To
Hello Lalit,
This question sounds more nearer to Shell scripting than Perl. No offence,
you can do this in Perl too, but since underlaying test scripts are shell
based (which you mentioned as 'i have implemented 50 shell program to test'
), i would suggest you to write a wrapper shell script which
On 09/09/2013 11:00, Hans Ginzel wrote:
Is there a shorter way to write $a = ! $a, please?
perl -le '
my @a = (undef, @ARGV);
for $a (@a) {
my @r; push @r, "$a:";
for (1..3) {
push @r, $a^=1; # <--
}
print "@r";
}
' 0 1 2 3 -1
: 1 0 1
0: 1 0 1
1: 0 1 0
2: 3 2 3
3:
On Mon, 09 Sep 2013 11:00:31 +0200
Hans Ginzel wrote:
> Is there a shorter way to write $a = ! $a, please?
Why?
--
Don't stop where the ink does.
Shawn
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On Tue, Sep 10, 2013 at 01:25:22AM -0700, Lalit Deshmukh wrote:
> i have implemented 50 shell program to test. now i want to run all these
> shell program one by one(in short test automation) (once first program
> getting executed then next will start). please let me know how to do this?
> in UNIX
On Sep 10, 2013, at 1:25 AM, Lalit Deshmukh wrote:
> Hi,
>
> i have implemented 50 shell program to test. now i want to run all these
> shell program one by one(in short test automation) (once first program
> getting executed then next will start). please let me know how to do this? in
> UNIX
On Tue, Sep 10, 2013 at 08:29:00AM -0400, Shawn H Corey wrote:
> On Mon, 09 Sep 2013 11:00:31 +0200
> Hans Ginzel wrote:
>
> > Is there a shorter way to write $a = ! $a, please?
>
> Why?
Because it's the sort of thing one might expect Perl to provide. I know
that I have thought that before.
^
On 09/10/13 14:59, David Christensen wrote:
Assuming canonical boolean values, post-invert semantics (save the new
value into another variable) ...
Pre-invert semantics (save the old value into another variable) ...
Oops -- it looks like I got my pre- and post- backwards...
And, dropping the
September 10, 2013 06:15 Hans Ginzel wrote:
> Is there a shorter way to write $a = ! $a, please?
> Something analogous to ++ and -- operators like $a !! or !! $a would
> negate the variable $a and return its previous or new value
> respectively.
I don't believe Perl has boolean pre-invert or post
On Tue, 10 Sep 2013 17:04:07 -0700
David Christensen wrote:
> scalar ($v ^= 1, !$v)
( ! ( $v ^= 1 ))
or
do { $v = ! $v; !$v }
TIM TOW TDI ;)
--
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Shawn
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beginners:
Here's a second try at using bitwise xor-equals to implement boolean
pre- and post-invert operations for variables containing canonical
boolean values (undef, empty string, zero, and one).
The pre-invert semantics case (invert, then use) uses the bitwise
xor-equals operator and o
On Mon, Sep 9, 2013 at 6:00 AM, Hans Ginzel wrote:
> Hello!
>
> Is there a shorter way to write $a = ! $a, please?
>
> Something analogous to ++ and -- operators like $a !! or !! $a would negate
> the variable $a and return its previous or new value respectively.
>
> Best regards
Not really. Bu
On Mon, Sep 9, 2013 at 5:00 AM, Hans Ginzel wrote:
> Hello!
>
> Is there a shorter way to write $a = ! $a, please?
>
> Something analogous to ++ and -- operators like $a !! or !! $a would negate
> the variable $a and return its previous or new value respectively.
>
It sounds like what you're req
David Christensen wrote:
September 10, 2013 06:15 Hans Ginzel wrote:
> Is there a shorter way to write $a = ! $a, please?
> Something analogous to ++ and -- operators like $a !! or !! $a would
> negate the variable $a and return its previous or new value
> respectively.
I don't believe Perl
On 09/10/13 20:01, John W. Krahn wrote:
xor-equals IS assignment and has the same precedence as assignment:
Thanks!
David
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