"printf -v" inside function not working with redirected output

Question

With bash 4.1.2 and 4.3.48, the following script gives the expected output:

#!/bin/bash

returnSimple() {
   local  __resultvar=$1
   printf -v "$__resultvar" '%s' "ERROR"
   echo "Hello World"
}

returnSimple theResult
echo ${theResult}
echo Done.

Output as expected:

$ ./returnSimple
Hello World
ERROR
Done.

However, when stdout from the function is piped to another process, the assignment of the __resultvar variable does not work anymore:

#!/bin/bash

returnSimple() {
   local  __resultvar=$1
   printf -v "$__resultvar" '%s' "ERROR"
   echo "Hello World"
}

returnSimple theResult | cat
echo ${theResult}
echo Done.

Unexpected Output:

$ ./returnSimple
Hello World

Done.

Why does printf -v not work in the second case? Should printf -v not write the value into the result variable independent of whether the output of the function is piped to another process?

@chepner Right, and that is also the reason why I observed that this use case does work with redirection, but not with piping ... fixed that detail in the question — Andreas Fester
– Andreas Fester, Commented Oct 12, 2017 at 13:45

randomir · Accepted Answer · 2017-10-12 13:31:26Z

8

See man bash, section on Pipelines:

Each command in a pipeline is executed as a separate process (i.e., in a subshell).

That's why when you write cmd | cat, cmd receives a copy of variable that it can't modify.

A simple demo:

$ test() ((a++))
$ echo $a

$ test
$ echo $a
1
$ test | cat
$ echo $a
1

answered Oct 12, 2017 at 13:31

randomir

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1 Comment

Andreas Fester Over a year ago

Thanks, that is exactly what I missed ... I just also found that out ;-)

Andreas Fester · Accepted Answer · 2017-10-12 13:33:11Z

Interestingly enough, the same also happens when using eval $__resultvar="'ERROR'" instead of the printf -v statement. Thus, this is not a printf related issue.

Instead, adding a echo $BASH_SUBSHELL to both the main script and the function shows that the shell spawns a sub shell in the second case - since it needs to pipe the output from the function to another process. Hence the function runs in a sub shell:

#!/bin/bash

returnSimple() {
    local  __resultvar=$1
    echo "Sub shell level: $BASH_SUBSHELL"
    printf -v "$__resultvar" '%s' "ERROR"
}

echo "Sub shell level: $BASH_SUBSHELL"
returnSimple theResult | cat
echo ${theResult}
echo Done.

Output:

% ./returnSimple.sh
Sub shell level: 0
Sub shell level: 1

Done.

This is the reason why any variable assignments from within the function are not passed back to the calling script.

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