topic Re: arithmetic operations problem in Operating System - HP-UX

arithmetic operations problem

Paul_481 — Wed, 01 Jun 2005 04:41:34 GMT

Hi,

This is using POSIX shell.

I.

#((x=1000000000+1));echo $x
1000000001

II.

#((x=10000000000+1));echo $x
1410065409

I notice that POSIX is not capable of computing more that 10 digit. Is there a way around this? Any other alternative?

Regards,
Paul

Re: arithmetic operations problem

Eric Antunes — Wed, 01 Jun 2005 04:58:04 GMT

Hi Paul,

Try this:

((x=500000000+1));((y=500000000));echo $x;echo $y;((z=$y+$x));echo $z

Best Regards,

Eric Antunes

Re: arithmetic operations problem

Paul_481 — Wed, 01 Jun 2005 05:26:27 GMT

Hi Eric,

((x=500000000+1));((y=500000000));echo $x;echo $y;((z=$y+$x));echo $z

From you example the value of x is still less than 10.

What I need is for a way to compute 11 digit integer.

Regards,
Paul

Re: arithmetic operations problem

CAS_2 — Wed, 01 Jun 2005 05:34:19 GMT

Arithmetic operations are 32 bit wide.

integers range are: from -2^31 (-2147483648) to +2^31-1 (+2147483647)

Following commands show only low 32-bit are used:

(( x=2147483647)); (( x=x+1 )); echo $x
-2147483648

I should use "bc" command (or "dc" utility) instead.

Re: arithmetic operations problem

erics_1 — Wed, 01 Jun 2005 05:39:16 GMT

Paul,

Take a look at Knowledge Base Doc. ID 4000099480 . This explains workarounds to 32 bit integer limitations.

Regards,
Eric

Re: arithmetic operations problem

CAS_2 — Wed, 01 Jun 2005 05:44:05 GMT

Using bc command:

echo 10000000000+1 | bc | read x
echo $x

Note that the 32-bit arithmetic is also applicable to INTEGER shell variables defined by means of 'typeset -i' (or 'integer' alias). Therefore, the shell variables used to handle this big values HAVE TO BE NON-INTEGER.

The following examples:

typeset -i x
echo 10000000000+1 | bc | read x
echo $x

typeset +i x
echo 10000000000+1 | bc | read x
echo $x

wil print:

1410065409
10000000001

Re: arithmetic operations problem

Eric Antunes — Wed, 01 Jun 2005 06:00:25 GMT

Sorry Paul, I missed one zero...

I also have this 32 bits maximum:

2^0+2^1+...+2^31=2^32-2^0=4294967296-1=4294967295 but since we also need to use negative numbers, our limit is as CAS said...

Best Regards,

Eric Antunes

Re: arithmetic operations problem

Paul_481 — Fri, 03 Jun 2005 02:35:33 GMT

Hi all,

I have problem storing the output of bc into variable. I need to get to correct results and process it further.

cscpcc@idmp005 [/home/cscpcc]
#print "2124582832+256000000"|bc
2380582832

cscpcc@idmp005 [/home/cscpcc]
#print "2124582832+256000000"|bc|read x|echo $x
-1914384464

Regards,
Paul

Re: arithmetic operations problem

Stephen Keane — Fri, 03 Jun 2005 02:47:17 GMT

You probably meant

# print "2124582832+256000000"|bc|read x;echo $x

(i.e. lose the last pipe)

Works for me!

Re: arithmetic operations problem

Stephen Keane — Fri, 03 Jun 2005 02:48:38 GMT

(well apart from the 32-bit integer rounding bit!)

Re: arithmetic operations problem

Paul_481 — Mon, 06 Jun 2005 08:04:51 GMT

Hello All,

Well I finally able to make my script work. Actually the script is for disk utilization computation from bdf output. My script is working ok until we upgrade our disk to EVA when we add a couple of Terbytes disk.

Any way, I used "bc" as suggested and it works.

Thanks everyone for the reply.

Regards,
Paul