Unix Arithmatic operation issue , datatype issue


 
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# 1  
Old 02-12-2008
Question Unix Arithmatic operation issue , datatype issue

Hi,
I have a shell scripting. This will take 7 digit number in each line and add 7 digit number with next subsequent lines ( normal addition ).

Eg:

0000001
0000220
0001235
0000022
0000023
...........
.........
........


Like this i am having around 1500000 records. After adding , I am getting the result 2147483647 but actual result is 2156379608 . I found the root cause of this issue is that Unix temproary variable can hold only 2156379608 ( Range of integer ) as it's 32 bit. For eg:

a=2147483647
expr $a + 2

you will get the negative result. Because while expr working, the result will be stored in internal temporary variable or register then you will get the result.. but that particular temporary variable or register can accomodate only 2147483647 .. if it crosses this limit, you may get the junk value like -ve values... this is my finiding for this issue.. But I want to have resolution for this issue. how to add or do arthimatic opration if i want to have result more than limit ( 2147483647 ). I mean

a=2147483647
expr $a + 2

for this i need to get the right result rather than -ve value.
# 2  
Old 02-12-2008
This will handle up to 15 signifcant digits:
Code:
 awk ' {total+=$0} END { print total} ' file

This will handle bigger numbers depending on your version of bc:
Code:
#/bin/ksh
total=0
while read value
do
    total=$(echo "$value + $total" | bc )
done < file
echo $total

# 3  
Old 02-13-2008
Question Reply

Hi ,

Thanks. We are also using the awk for arthitmatic operation . But we are not getting exepected result. I tried with your code as well.

$ awk ' {total+=$0} END { print total} ' hash_column_wrong.txt
2.15638e+09

I got the result in expontial format. But I need the result with all the digits. How to achive this?. Then second method which you shown using the bc calculator is taking more time to complte. So I think awk is the best method. As of now , I am getting result as 2.15638e+09 . but I need to get result like 2156380000 . Please help me.

Regards,
Thambi
# 4  
Old 02-13-2008
can you please provide the list of hash_column_wrong.txt ????
# 5  
Old 02-13-2008
Reply..

As I already mentioned, it just contains the 7 digiti number in every row. Like these, this file has around 150000 . We just need to add all the rows..that's all

0000001
0000343
0000001
0001426
0000001
0000718
0000162
0000103
0000021
0000011
0000016
0000312
0000026
0000031
0000005
0000022
0000001
0000001
0011845
0003713
0000001
0000011
0000001
0000001
0000001
..,.,
........
.......
# 6  
Old 02-13-2008
Use awk to prepare the file for bc.

{ awk '{ print $0 " +\\" }' bigfile; echo 0; } | bc

Assuming bc is using the arbitrary precision library, you shouldn't have a problem here.
# 7  
Old 02-13-2008
Reply....

When I run the below command

awk '{ print $0 " +\\" }' hash_column_wrong.txt | bc

I got below exception..

bundling space exceeded on line 1,
bundling space exceeded on line 1,
bundling space exceeded on line 1,

is it due to \n character that occurs end of the line?.
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