Home » How can I get bash to exit on backtick failure in a similar way to pipefail?

How can I get bash to exit on backtick failure in a similar way to pipefail?

Solutons:


The exact language used in the Single UNIX specification to describe the meaning of set -e is:

When this option is on, if a simple command fails for any of the reasons listed in Consequences of Shell Errors or returns an exit status value >0, and is not [a conditional or negated command], then the shell shall immediately exit.

There is an ambiguity as to what happens when such a command occurs in a subshell. From a practical point of view, all the subshell can do is exit and return a nonzero status to the parent shell. Whether the parent shell will in turn exit depends on whether this nonzero status translates into a simple command failing in the parent shell.

One such problematic case is the one you encountered: a nonzero return status from a command substitution. Since this status is ignored, it does not cause the parent shell to exit. As you’ve already discovered, a way to take the exit status into account is to use the command substitution in a simple assignment: then the exit status of the assignment is the exit status of the last command substitution in the assignment(s).

Note that this will perform as intended only if there is a single command substitution, as only the last substitution’s status is taken into account. For example, the following command is successful (both according to the standard and in every implementation I’ve seen):

a=$(false)$(echo foo)

Another case to watch for is explicit subshells: (somecommand). According to the interpretation above, the subshell may return a nonzero status, but since this is not a simple command in the parent shell, the parent shell should continue. In fact, all the shells I know of do make the parent return at this point. While this is useful in many cases such as (cd /some/dir && somecommand) where the parentheses are used to keep an operation such as a current directory change local, it violates the specification if set -e is turned off in the subshell, or if the subshell returns a nonzero status in a way that would not terminate it, such as using ! on a true command. For example, all of ash, bash, pdksh, ksh93 and zsh exit without displaying foo on the following examples:

set -e; (set +e; false); echo "This should be displayed"
set -e; (! true); echo "This should be displayed"

Yet no simple command has failed while set -e was in effect!

A third problematic case is elements in a nontrivial pipeline. In practice, all shells ignore failures of the elements of the pipeline other than the last one, and exhibit one of two behaviors regarding the last pipeline element:

  • ATT ksh and zsh, which execute the last element of the pipeline in the parent shell, do business as usual: if a simple command fails in the last element of the pipeline, the shell executing that command, which happens to be the parent shell, exits.
  • Other shells approximate the behavior by exiting if the last element of the pipeline returns a nonzero status.

Like before, turning off set -e or using a negation in the last element of the pipeline causes it to return a nonzero status in a way that should not terminate the shell; shells other than ATT ksh and zsh will then exit.

Bash’s pipefail option causes a pipeline to exit immediately under set -e if any of its elements returns a nonzero status.

Note that as a further complication, bash turns off set -e in subshells unless it’s in POSIX mode (set -o posix or have POSIXLY_CORRECT in the environment when bash starts).

All of this shows that the POSIX specification unfortunately does a poor job at specifying the -e option. Fortunately, existing shells are mostly consistent in their behavior.

(Answering my own because I’ve found a solution)
One solution is to always assign this to an intermediate variable.
This way the returncode ($?) is set.

So

ABC=`exit 1`
echo $?

Will output 1 (or instead exit if set -e is present), however:

echo `exit 1`
echo $?

Will output 0 after a blank line. The return code of the echo (or other command that ran with the backtick output) will replace the 0 return code.

I am still open to solutions that do not require the intermediate variable, but this gets me some of the way.

As the OP pointed out in his own answer, assigning the output of the subcommand to a variable does solve the problem; the $? is left unscathed.

However, one edge case can still puzzle you with false negatives (i.e. command fails but error doesn’t bubble up), local variable declaration:

local myvar=$(subcommand) will always return 0!

bash(1) points this out:

   local [option] [name[=value] ...]
          ... The return status is 0 unless local is used outside a function,
          an invalid name is supplied, or name is a readonly variable.

Here’s a simple test case:

#!/bin/bash

function test1() {
  data1=$(false) # undeclared variable
  echo 'data1=$(false):' "$?"
  local data2=$(false) # declaring and assigning in one go
  echo 'local data2=$(false):' "$?"
  local data3
  data3=$(false) # assigning a declared variable
  echo 'local data3; data3=$(false):' "$?"
}

test1

The output:

data1=$(false): 1
local data2=$(false): 0
local data3; data3=$(false): 1

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