How can I add numbers in a Bash script?

Question

I have this Bash script and I had a problem in line 16. How can I take the previous result of line 15 and add it to the variable in line 16? ``` #!/bin/bash num=0 metab=0 for ((i=1; i<=2; i++)); do for j in `ls output-$i-*`; do echo "$j" metab=$(cat $j|grep EndBuffer|awk '{sum+=$2} END { print sum/120}') (line15) num= $num + $metab (line16) done echo "$num" done ```

Accepted Answer

For integers:

Use arithmetic expansion: `$((EXPR))`

```
num=$((num1 + num2))
num=$(($num1 + $num2))       # Also works
num=$((num1 + 2 + 3))        # ...
num=$[num1+num2]             # Old, deprecated arithmetic expression syntax
```

Using the external `expr` utility. Note that this is only needed for really old systems.

```
num=`expr $num1 + $num2`     # Whitespace for expr is important
```

For floating point:

Bash doesn't directly support this, but there are a couple of external tools you can use:

```
num=$(awk "BEGIN {print $num1+$num2; exit}")
num=$(python -c "print $num1+$num2")
num=$(perl -e "print $num1+$num2")
num=$(echo $num1 + $num2 | bc)   # Whitespace for echo is important
```

You can also use scientific notation (for example, `2.5e+2`).

Common pitfalls:

When setting a variable, you cannot have whitespace on either side of `=`, otherwise it will force the shell to interpret the first word as the name of the application to run (for example, `num=` or `num`)

`num= 1` `num =2`

`bc` and `expr` expect each number and operator as a separate argument, so whitespace is important. They cannot process arguments like `3+` `+4`.

`num=`expr $num1+ $num2``

How can I add numbers in a Bash script?

Solution

Problem