float numbers division bash

Question

I have read all previous post about float numbers divisions in bash but I can't solve my problem...

I have this bash script:

lengthseq=$(sed -e 's/^\(>\).*//' sequence.fasta | sed '1d' | tr -d "\n" | awk '{ print length }')
lengthcodons=$("$lengthseq/3" | bc -l)

echo $lengthseq
echo $lengthcodons

Lengthseq it's ok. echo $lengthseq prints 2275859.

If I run it in konsole:

echo "2275859/3" | bc -l
758619.66666666666666666666

It's okay too.

So I don't understant why if I try it in my script, $lengthcodons generates and error.

Could I ask you want the command `lengthseq=$(sed -e 's/^$>$.*//' sequence.fasta | sed '1d' | tr -d "\n" | awk '{ print length }')` does? (in fact, I'd like to know if it's really doing what you're expecting). — gniourf_gniourf, Nov 10 '13 at 16:31
Possible duplicate of [How do I use floating-point division in bash?](https://stackoverflow.com/questions/12722095/how-do-i-use-floating-point-division-in-bash) — Andrea Corbellini, Apr 29 '18 at 02:05

score 3 · Accepted Answer · answered Nov 10 '13 at 15:42

3

Correct syntax of using bc -l is:

lengthcodons=$(bc -l <<< "$lengthseq/3")
echo "$lengthcodons"
758619.66666666666666666666

Or with scale=2

lengthcodons=$(bc -l <<< "scale=2; $lengthseq/3")
echo "$lengthcodons"
758619.66

answered Nov 10 '13 at 15:42

anubhava

761,203
64
569
643

score 2 · Answer 2 · answered Nov 10 '13 at 15:45

2

You're missing echo in following line

lengthcodons=$(echo "$lengthseq/3" | bc -l)

answered Nov 10 '13 at 15:45

jkshah

11,387
6
35
45

score 1 · Answer 3 · answered Nov 10 '13 at 15:45

1

Here:

lengthcodons=$("$lengthseq/3" | bc -l)

You're executing the line "2275859/3" | bc -l and put the result in lengthcodons, your forgot the echo.

answered Nov 10 '13 at 15:45

Maxime Chéramy

17,761
8
54
75

float numbers division bash

3 Answers3