Passing several values through a pipe in bash

I have been fiddling with some git-related shell scripts, and decided to try and follow the same approach as git in their structure. This means using the Unix system where each piece of functionality is a separate script (or executable) that communicates by using command-line arguments, reading from the standard input stream, and writing output to the standard output stream.

This allows each piece of functionality to be written in any programming or scripting language. In git's case this has allowed initial versions to be written in bash or perl and later optimised versions (sometimes written in C) to be dropped in, piece by piece. It's an incredibily flexible way of working and can also be very efficient.

Most of my prototyping has been in bash, and I've found sometimes I need to write out multiple values from a script and collect them as input in another script.

Writing the output is simple:

#!/bin/bash

# outputter.bash

# Imagine A, B and C have been created by some complex process:
A="foo bar"
B="  bar"
C="baz   "

# At the end of our script we simply write them out on separate lines in a known order
echo "${A}"
echo "${B}"
echo "${C}"

But reading them in somewhere else gave me some trouble until I learned this recipe:

#!/bin/bash

# inputter.bash

# Read in the values one per line:
IFS=$'\n' read A
IFS=$'\n' read B
IFS=$'\n' read C

# Now we can use them.
echo "A='${A}'"
echo "B='${B}'"
echo "C='${C}'"

And now the values transfer succesfully, preserving whitespace:

$ ./outputter.bash | ./inputter.bash
A='foo bar'
B='  bar'
C='baz   '

The recipe uses bash's built-in read command to populate the variables, but sets the IFS variable (Internal Field Separator) to a newline, meaning all the whitespace in the line is treated as part of the value to be read. The $'\n' syntax is a literal newline.