CODEWITHSUNDEEP
bashshellunixsh
Tricky Dicky
Tricky Dicky

Reputation: 95

Bash floating points and negative values

Hi i am tring to produce flaoting point precision using the following code

let number1=0 number2=0 operator=+

printf "%0.2f\n" result=$(( number1 $operator number2 ))

The code works without the printf but i cannot figure out how to perform negative(-) calcs and floating points?

Upvotes: 2

Views: 3645

Answers (2)

Tricky Dicky
Tricky Dicky

Reputation: 95

In the end i figured it out!

result=$(echo "scale=4; (( $number1 $operator $number2 ))" | bc)

Upvotes: 0

Rubens
Rubens

Reputation: 14778

Bash does not support floating point calculations, so, either you multiply the numbers being operated by as many zeros as decimals you want:

# 10.321 - 123.01
result=$(( 10321 - 123010 ))
echo ${result:0:-3}.${result:${#result} - 3}

Or simply use another tool to do this, like bc:

echo "scale=2; 10.321 - 123.01" | bc

Also, the syntax you've used is not valid; you should have:

printf "%0.2f\n" $(( number1 $operator number2 ))

Upvotes: 2

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