Shifting operation on Perl 5.26 does not work as intended

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Hello All,

I have a below piece of code:

$tempVar = hex(get_data());

$test |= $tempVar << (64 - $shiftby * ($size - $iteration - 1));

This code is to be executed on perl 5.26. Currently it is executing on 5.16.

When i try to execute this on 5.26, it gives random results (Mostly 0 everytime). It works fine on 5.16.

I needed some help on how to handle this on 5.26. Basically the below equation should be modified to run on both 5.16 and 5.26:

$test |= $tempVar << (64 - $shiftby * ($size - $iteration - 1));

Any help/guidance would be appreciated.

Thanks

What I have tried:

I tried the below:

1) Went through the perl config operations which are used to build perl 5.26 and perl 5.16.

2) Perl 5.26 is built with "

Quote:
-Duse64bitint

". I think due to this it explicitly treats all variables as 64 bits. However Perl 5.16 is not built with this.

3) I am trying to build perl 5.26 without this flag. But this will be cumbersome as we might move to some other version in near future. I am not sure if this is the right approach.

4) Finding alternative of this for example:

$tempVar |= $tempVar << (64 - $shiftby * ($size - $iteration - 1))
$test |= $tempVar

This however gives some defined value but with some delta.

Posted 14-May-24 5:38am

Rahul VB

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k5054 · Answer 1 · 2024-05-14T07:00:00

Solution 1

At a guess, you're experiencing a difference in how negative shifts are performed. Given

PERL

$t = 256;
$v = $t << -1;
print ("v = $v\n");

Running this on perl 5.16 (CentOS 7), I get 0, but using perl 5.36 (Debian bookworm), I get a value of 128.
If 64 - $shiftby * ($size -$iteration -1) is negative, you'll need to decide what you should do. Maybe right-shift instead? Abort? other?

Posted 14-May-24 7:00am

k5054

Comments

Rahul VB 15-May-24 5:09am

Thanks for the quick response. But, i dont get negetive values. I am exploring more.

k5054 15-May-24 12:16pm

Only thing I can suggest is that you build up your expression a bit at a time and examine the values as you go e.g.

$x = ($size - iteration -1);
$y = $shiftby * $x;
$z = 64 - $shiftby;
$val  = $TestVal << $z;
$test |= $val;
print STDERR, "x = $x, y = $y, z = $z, val = $val, test = $test\n';

At least then, you might get an idea where the two versions diverge.

Rahul VB 22-May-24 12:30pm

Thanks, for the response. I had to correct the shift logic which was implemented after breaking the code further down.