How to round or convert a float value to int with bc? getting: "(standard_in) 1: syntax error"

Question

I will get value like 2743410360.320 and I want value like 2743410360 to a variable.

I tried

INTValueOfGB=$ echo "($gb+0.5)/1" | bc

But I am getting (standard_in) 1: syntax error

Please check that your code's syntax is still the one you intended to post after I edited it. (That solitary `$` sign looks quite interesting.) — manatwork, Nov 15 '13 at 11:23

score 10 · Answer 1 · answered Nov 15 '13 at 11:36

10

You can use printf for rounding:

$ printf "%.0f" 2743410360.320

answered Nov 15 '13 at 11:36

dchirikov

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In `bash` 4.1 the builtin `printf` says “invalid number”. Works fine with `/usr/bin/printf`. – manatwork Nov 15 '13 at 11:39
@manatwork Works for me with bash 4.1.5 on Debian squeeze and even bash 3.2.39 on Ubuntu 8.04. – Gilles 'SO- stop being evil' Nov 16 '13 at 22:18
Indeed. Works fine in `bash` 4.2. Or maybe the problem was not the version – the first try was on Cygwin. – manatwork Nov 17 '13 at 12:36
I tried it is not working fine when i am passing a value $ printf "%.0f" $gb – vin Nov 18 '13 at 05:58
@manatwork: Are you using `%i` or `%.0f`? The former says 'invalid number' for me, the latter works. – Flimzy Mar 06 '15 at 15:14
@Flimzy, I am quite (though not 100%) sure it was `%.0f`. My bad, I not specified the patch level of the version. It works fine in `bash` 4.1.5, the oldest I currently have. – manatwork Mar 06 '15 at 15:59

score 7 · Answer 2 · answered Nov 01 '15 at 20:32

I don't see an answer to @vin's problem, which is:

But I am getting (standard_in) 1: syntax error

The bc command prints (standard_in) 1: syntax error because the shell variable gb is not set:

$ unset gb
$ echo "($gb+0.5)/1" | bc
(standard_in) 1: syntax error

$ gb=2743410360.320
$ echo "($gb+0.5)/1" | bc
2743410360

In the comments to the @dchirikov's answer, @vin says printf "%.0f" is "not working":

$ unset gb
$ printf '%.0f\n' $gb
0

$ gb=2743410360.320
$ printf '%.0f\n' $gb
2743410360

In both of the areas where @vin has problems, the unset variable reproduces the problem, and setting the variable solves the problem.

score 4 · Answer 3 · answered Nov 15 '13 at 11:26

4

$ p=2743410360.320
$ echo $p
2743410360.320
$ echo ${p%%.*}
2743410360

answered Nov 15 '13 at 11:26

michas

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The `+0.5` in the original code looks like _rounding_ is needed. – manatwork Nov 15 '13 at 11:36
simple - effective code – 3kstc Aug 08 '16 at 04:04
1

5.9 rounded is not 5 This doesn't round at all. Its chopping off everything after the decimal. – Inetquestion Jul 16 '21 at 18:14

score 3 · Answer 4 · answered Nov 01 '15 at 19:28

Your command needs a couple of ( ):

INTValueOfGB=$( echo "($gb+0.5)/1" | bc )

But that will not round the number, for that to work you need to set scale to 0:

INTValueOfGB=$( echo "scale=0;($gb+0.5)/1" | bc )

That will round up (from x.5) to the next integer.

To get "round to next even integer" (Banker's rule), you need to use printf rounding:

LC_NUMERIC=C printf "%.0f" 2743410360.320

The LC_NUMERIC is to ensure that numbers are processed with a dot as decimal. That is not what users in Germany will like to use, this should work with decimals with ,:

$ LC_NUMERIC=de_DE.utf8 printf "%.0f" 2743410360,320
2743410360

How to round or convert a float value to int with bc? getting: "(standard_in) 1: syntax error"

4 Answers4