int() function in awk changes the argument even if this is integer already

Question

I don't understand the following two results in awk

$ awk 'BEGIN {
  print    (log(.01)/log(10))
  print int(log(.01)/log(10))
}'
-2
-1

Version:

$ awk --version 
GNU Awk 5.0.1, API: 2.0 (GNU MPFR 4.0.2, GNU MP 6.1.2)

even though the mathematical result is an integer, the value `log(.01)` is for sure a real as it represents the natural logarithm. On top of that, there are floating-point approximations which make `log(0.01)` to be slightly different numerically then mathematically. Dito for `log(10)`. Ergo, the division will not be perfect. — kvantour, Oct 10 '19 at 09:35
Possible duplicate of [Is floating point math broken?](https://stackoverflow.com/questions/588004/is-floating-point-math-broken) — kvantour, Oct 10 '19 at 09:35

score 4 · Accepted Answer · answered Oct 09 '19 at 19:53

4

There is more than meets the eye:

$ awk 'BEGIN {
  printf "%.16f\n", log(.01)/log(10)
}'
-1.9999999999999996

answered Oct 09 '19 at 19:53

James Brown

So what format does `awk` use by default? `"%.6f\n"`? – Enlico Oct 09 '19 at 19:57
3

@EnricoMariaDeAngelis It's defined in the `OFMT` variable (https://www.gnu.org/software/gawk/manual/gawk.html#OFMT), which defaults to `%.6g`. – Benjamin W. Oct 09 '19 at 21:11

1 Answers1