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Accepted
Adam Matan
Asked: 2012-11-16 04:35:15 +0800 CST

bc: set number of digits after decimal point

  • 772

bc handles numbers as integers:

# echo "100/3" | bc
33

bc -l handles numbers as floating point objects:

# echo "100/3" | bc -l
33.33333333333333333333

Is there a way to limit the number of digits after the decimal point?

command-line

5 Answers

  1. Best Answer

    Set the scale special variable:

    $ echo "scale=2; 100/3" | bc
33.33
    • 113

  2. scale works only for division; if some geeks need it in multiplication, then you can do achieve this by using string manipulation. Say if you need to multiply 32 * 0.60 , answer is 19.20. If you need to get it 19 alone in answer you can get it by different methods.

    1. Using String Manipulation 

      $ S=$(echo "32*.60" | bc ) ; echo ${S%.*}
19

      String Manipulation syntax: ${Variable%pattern}, this will delete short matching pattern that comes after %. For more String manipulation details see the Advanced Bash-Scripting Guide.

    2. Using Scale as stated by **chronitis**

      $ echo "scale=0; 32*60/100" | bc
19

    3. To get rid of the trailing 0s, instead of string manipulation, one can also do a divide by 1.

      $ echo "0.232 * 1000" | bc
232.000

$ echo "0.232 * 1000 / 1" | bc
232
    • 20

  3. In addition to previous answers

    echo "scale=2; 1.0150876" | bc

    Returns

    1.0150876

    Add Math operations to get only 2 decimal numbers - (NUMBER*100)/100

    echo "scale=2; (1.0150876 * 100) / 100" | bc

    Now returns

    1.01
    • 6

  4. you can also use printf command to round off result upto 3 decimals 

    # printf "%.3f\n" $(echo "100/3" | bc -l)
3.333
    • 6

  5. Round-off

    scale=2 truncates the answer to two decimal digits, but we can achieve round-off like so:

    $ echo "a=12/104; scale=2; (a+0.005)/1" | bc -l
.12
    • 3