Printing long integers in awk

Question

I have a pipe delimited feed file which has several fields. Since I only need a few, I thought of using awk to capture them for my testing purposes. However, I noticed that printf changes the value if I use "%d". It works fine if I use "%s".

Feed File Sample:

[jaypal:~/Temp] cat temp

302610004125074|19769904399993903|30|15|2012-01-13 17:20:02.346000|2012-01-13 17:20:03.307000|E072AE4B|587244|316|13|GSM|1|SUCC|0|1|255|2|2|0|213|2|0|6|0|0|0|0|0|10|16473840051|30|302610|235|250|0|7|0|0|0|0|0|10|54320058002|906|722310|2|0||0|BELL MOBILITY CELLULAR, INC|BELL MOBILITY CELLULAR, INC|Bell Mobility|AMX ARGENTINA SA.|Claro aka CTI Movil|CAN|ARG|

I am interested in capturing the second column which is 19769904399993903.

Here are my tests:

[jaypal:~/Temp] awk -F"|" '{printf ("%d\n",$2)}' temp
19769904399993904   # Value is changed

However, the following two tests works fine -

[jaypal:~/Temp] awk -F"|" '{printf ("%s\n",$2)}' temp
19769904399993903   # Value remains same

[jaypal:~/Temp] awk -F"|" '{print $2}' temp
19769904399993903   # Value remains same

So is this a limit of "%d" of not able to handle long integers. If thats the case why would it add one to the number instead of may be truncating it?

I have tried this with BSD and GNU versions of awk.

Version Info:

[jaypal:~/Temp] gawk --version
GNU Awk 4.0.0
Copyright (C) 1989, 1991-2011 Free Software Foundation.

[jaypal:~/Temp] awk --version
awk version 20070501

Answer 1

I believe the underlying numeric format in this case is an IEEE double. So the changed value is a result of floating point precision errors. If it is actually necessary to treat the large values as numerics and to maintain accurate precision, it might be better to use something like Perl, Ruby, or Python which have the capabilities (maybe via extensions) to handle arbitrary-precision arithmetic.

Answer 2

Starting with GNU awk 4.1 you can use --bignum or -M

$ awk 'BEGIN {print 19769904399993903}'
19769904399993904

$ awk --bignum 'BEGIN {print 19769904399993903}'
19769904399993903

§ Command-Line Options

Answer 3

UPDATE: Recent versions of GNU awk support arbitrary precision arithmetic. See the GNU awk manual for more info.

ORIGINAL POST CONTENT: XMLgawk supports arbitrary precision arithmetic on floating-point numbers. So, if installing xgawk is an option:

zsh-4.3.11[drado]% awk --version |head -1; xgawk --version | head -1
GNU Awk 4.0.0
Extensible GNU Awk 3.1.6 (build 20080101) with dynamic loading, and with statically-linked extensions

zsh-4.3.11[drado]% awk 'BEGIN {
  x=665857
  y=470832
  print x^4 - 4 * y^4 - 4 * y^2
  }'
11885568

zsh-4.3.11[drado]% xgawk -lmpfr 'BEGIN {
  MPFR_PRECISION = 80
  x=665857
  y=470832
  print mpfr_sub(mpfr_sub(mpfr_pow(x, 4), mpfr_mul(4, mpfr_pow(y, 4))), 4 * y^2)
  }'
1.0000000000000000000000000

Answer 4

This answer was partially answered by @Mark Wilkins and @Dennis Williamson already but I found out the largest 64-bit integer that can be handled without losing precision is 2^53. Eg awk's reference page http://www.gnu.org/software/gawk/manual/gawk.html#Integer-Programming

(sorry if my answer is too old. Figured I'd still share for the next person before they spend too much time on this like I did)

Answer 5

You're running into Awk's Floating Point Representation Issues. I don't think you can find a work-around within awk framework to perform arithmetic on huge numbers accurately.

Only possible (and crude) way I can think of is to break the huge number into smaller chunk, perform your math and join them again or better yet use Perl/PHP/TCL/bsh etc scripting languages that are more powerful than awk.

Answer 6

Using nawk on Solaris 11, I convert the number to a string by adding (concatenate) a null to the end, and then use %15s as the format string:

printf("%15s\n", bignum "")