bash shell script which adds output of commands

Question

Let's say I have a command called foo which prints a number to the screen when called:

$ foo
3

Let's also say I have another command called bar which prints another number to the screen when called:

$ bar
5

I'm looking to write a shell script which will arithmetically add together the output of foo and bar (3+5=8). How would I do that? (The outputs from the commands are not known ahead of time. They just so happen to have been 3 and 5 the last time they were run. They could have been something else.)

Answer 1

Use bash's let to evalutate arithmetric expressions.

#!/bin/bash
a=`echo 3`
b=`echo 5`

let c=$a+$b
echo $c

Just substitute the calls to echo with your program calls.

Answer 2

An alternative to let is to use double-parenthesis syntax:

(( c = $(foo) + $(bar) ))

or

echo $(( $(foo) + $(bar) ))

or using variables, you can omit the dollar sign on the right hand side of the equals:

(( c += $(foo) + num ))

(which also illustrates an incremental assignment)

If you're using non-integers you can use bc:

echo "$(foo) + $(bar)" | bc

or

c=$(echo "$(foo) + $(bar)" | bc)

One advantage of using double parentheses is that you can put spaces around the operands and operators to make things more readable:

(( c = ( a + b ) * ( i - j ) ))

Answer 3

bash:

bc < <({ foo ; echo + ; bar ; } | tr '\n' ' ' ; echo)

If the output is integers only:

$(( $(foo) + $(bar) ))