How to print the first three digits after the decimal point of a floating number?

Question

I am new to BASH and learnt that both printf and echo write to the standard output. But, I came through the following example which reads an expression from input and computes the result to accuracy of 3 decimal places:

read exp
printf "%.3f\n" "$(echo $exp | bc -l)"

I don't get why echo is passed here as an argument in the the printf statement. Is there any other way to represent the statement using only echo?

Answer 1

After read exp, you can use echo as follow:

$ echo "scale=3; $exp/1" | bc
9.456

But note that this command does not require echo:

$ bc <<< "scale=3; $exp/1"
9.456

EDIT: in order to get a rounded result you have to use the printf bash builtin as bc can't round values.

The following command works as expected:

 echo "scale=4; $exp" | bc | xargs printf "%.3f\n"

Answer 2

Here is a solution using only echo:

• if you are using a decimal point (.) in your rational number:

  $ read exp
  123.4567
  $ int=${exp%%.*}
  $ rat=${exp##*.}
  $ echo $int.${rat:0:3}
  123.456
  

• if you are using a decimal comma (,) in your rational number (as I use and most part of continental Europe):

  $ read exp
  123,4567
  $ int=${exp%%,*}
  $ rat=${exp##*,}
  $ echo $int,${rat:0:3}
  123,456
  

For more info see Advanced Bash-Scripting Guide: Manipulating Strings.