Adding variables in bash

Question

I am trying to assign 2 variables floating values and then trying to store the sum in third variable but I get an error.

#!/bin/bash
x=0.1
y=1000.0
z=$((x+y))

echo $z

output is: xyz.sh: 6: xyz.sh: Illegal number: 0.1

I am unable to understand what is going wrong.

I want z to be = 1000.01

anubhava · Accepted Answer · 2015-02-02 09:38:09Z

3

BASH doesn't support floating point arithmetic. Use bc command instead:

z=$(bc -l <<< "$x + $y")
echo "$z"
1000.1

answered Feb 2, 2015 at 9:38

anubhava

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4 Comments

Sneha Sharma Over a year ago

Thanks. I tried using bc and it worked out. #!/bin/bash x=0.1 y=1000.0 z= echo "$x+$y"|bc echo "$z"

Sneha Sharma Over a year ago

I got stuck again. its actually not storing value in z but just printing it because i used z= echo "$x+$y"|bc when I used z=$(bc -l <<< "$x + $y") it says redirectionunexpected

Sneha Sharma Over a year ago

using that syntax gives me a syntax error which say xyz.sh: 6: xyz.sh: Syntax error: redirection unexpected

anubhava Over a year ago

Can you show you full script by editing the question. Make sure you are using bash not sh

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