CODEWITHSUNDEEP
bashfunction
David W.
David W.

Reputation: 107040

bash way of setting multiple variables from a function

I want to write a shell script in a more top downish design. I've use to do this with Kornshell scripts. I define various functions that may return multiple set of variables (for example, a getopts function). I can then use these functions in my main program to set the variables I want.

Unfortunately, BASH and Kornshell seem to diverge in the way they handle this instance. Here's a simple script I run in the Kornshell. My function foo returns four values. I'll read in these four values into variables in my main program:

#!/bin/ksh

function foo {
    echo "this:that:the:other"
}

IFS=":"
foo | read one two three four
echo "one = $one"
echo "two = $two"
echo "three = $three"
echo "four = $four"

This produces:

one = this
two = that
three = the
four = other

Works like a charm. Let's try the same program using BASH instead of Kornshell:

#!/bin/bash

function foo {
    echo "this:that:the:other"
}

IFS=":"
foo | read one two three four
echo "one = $one"
echo "two = $two"
echo "three = $three"
echo "four = $four"

This produces:

one = 
two = 
three = 
four =

The pipe to the read doesn't work at all. Let's try this as a hereis read:

#!/bin/bash

function foo {
    echo "this:that:the:other"
}

IFS=":"
read one two three four<<<$(foo)
echo "one = $one"
echo "two = $two"
echo "three = $three"
echo "four = $four"

I get:

one = this that the other
two =
three =
four =

That allowed $one to be set, but not the other variables. Strangely, the colons were removed from the string. Let's remove the colons from the return value:

#!/bin/bash

function foo {
    echo "this that the other"
}

read  one two three four<<<$(foo)
echo "one = $one"
echo "two = $two"
echo "three = $three"
echo "four = $four"

one = this
two = that
three = the
four = other

That does work. Each variable is read in from function foo.

What am I doing wrong? Why doesn't setting IFS seem to work the way I would think it would in BASH? Or is there a better way to do this?

Upvotes: 1

Views: 6423

Answers (2)

anubhava
anubhava

Reputation: 785038

Without using read this will also work in both bash and ksh:

IFS=: && set -- `foo`
echo "one = $1"
echo "two = $2"
echo "three = $3"
echo "four = $4"

bts This also works in BASH without creating a sub-shell:

function foo {
    echo "this:that:the:other"
}

IFS=":" read one two three four < <(foo)
echo "one = $one"
echo "two = $two"
echo "three = $three"
echo "four = $four"

Upvotes: 1

fedorqui
fedorqui

Reputation: 289545

You have to quote:

#!/bin/bash

function foo {
    echo "this:that:the:other"
}

IFS=":"
read one two three four<<<"$(foo)"
                          ^      ^

When executed it returns:

$ ./a
one = this
two = that
three = the
four = other

Regarding this one not working:

#!/bin/bash

...    

IFS=":"
foo | read one two three four
echo "one = $one"

I guess it is because IFS is defined in a different shell than the one having read. Also, to pipe foo to read doesn't seem to be the way to give a string to read in bash.

To do it, open a new shell, as seen on Bash script, read values from stdin pipe:

$ foo | { IFS=":"; read one two three four; echo "one = $one"; echo "two = $two"; echo "three = $three"; echo "four = $four"; }
one = this
two = that
three = the
four = other

Upvotes: 6

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