Why does `if $(true) ; then ... fi` succeed?
Asked Answered
D

5

23

Inspired by this question:

What should an if statement do when the condition is a command substitution where the command produces no output?

NOTE: The example is if $(true); then ..., not if true ; then ...

For example, given:

if $(true) ; then echo yes ; else echo no ; fi

I would think that $(true) should be replaced by the output of the true command, which is nothing. It should then be equivalent to either this:

if "" ; then echo yes ; else echo no ; fi

which prints no because there is no command whose name is the empty string, or to this:

if ; then echo yes ; else echo no ; fi

which is a syntax error.

But experiment shows that if the command produces no output, the if statement treats it as true or false depending on the status of the command, rather than its output.

Here's a script that demonstrates the behavior:

#!/bin/bash

echo -n 'true:          ' ; if true          ; then echo yes ; else echo no ; fi
echo -n 'false:         ' ; if false         ; then echo yes ; else echo no ; fi
echo -n '$(echo true):  ' ; if $(echo true)  ; then echo yes ; else echo no ; fi
echo -n '$(echo false): ' ; if $(echo false) ; then echo yes ; else echo no ; fi
echo -n '$(true):       ' ; if $(true)       ; then echo yes ; else echo no ; fi
echo -n '$(false):      ' ; if $(false)      ; then echo yes ; else echo no ; fi
echo -n '"":            ' ; if ""            ; then echo yes ; else echo no ; fi
echo -n '(nothing):     ' ; if               ; then echo yes ; else echo no ; fi

and here's the output I get (Ubuntu 11.04, bash 4.2.8):

true:          yes
false:         no
$(echo true):  yes
$(echo false): no
$(true):       yes
$(false):      no
"":            ./foo.bash: line 9: : command not found
no
./foo.bash: line 10: syntax error near unexpected token `;'
./foo.bash: line 10: `echo -n '(nothing):     ' ; if               ; then echo yes ; else echo no ; fi'

The first four lines behave as I'd expect; the $(true) and $(false) lines are surprising.

Further experiment (not shown here) indicates that if the command between $( and ) produces output, its exit status doesn't affect the behavior of the if.

I see similar behavior (but different error messages in some cases) with bash, ksh, zsh, ash, and dash.

I see nothing in the bash documentation, or in the POSIX "Shell Command Language" specification, to explain this.

(Or perhaps I'm missing something obvious.)

EDIT : In light of the accepted answer, here's another example of the behavior:

command='' ; if $command ; then echo yes ; else echo no ; fi

or, equivalently:

command=   ; if $command ; then echo yes ; else echo no ; fi
Devil answered 22/1, 2012 at 23:59 Comment(4)
This is very interesting. Anyway, I think there's an error in your question. You said that if "" ; then echo yes ; else echo no ; fi returns no because 'there is no command whose name is the empty string', but by that same logic it should fail with an error something like "<empty string>: command not found". There is some other difference between if "" ; then echo yes ; else echo no ; fi and if "nonexistant" ; then echo yes ; else echo no ; fiLoriloria
@AaronMcDaid: I think there’s an error in your comment.  if "nonexistent" ; then … produces the error message bash: nonexistent: command not found and then says no.  By comparison, if "" ; then … produces the error message bash: : command not found and then says no.  That’s exactly the same behavior, except nonexistent has been deleted (i.e., replaced by nothing).  It is the same message that you get if you type just "" (followed by <Enter>).  Can you give an example of a case where bash says something like <empty string> instead of just writing an empty string?  … (Cont’d)Diabolo
(Cont’d) …  Can you identify any actual difference between if "" ; then … and if "nonexistent" ; then …?Diabolo
@KeithThompson: I agree that this is an interesting question.  With all the illuminating examples presented on this page, I am surprised that nobody has offered if "$(true)" ; then … and command='' ; if "$command" ; then … (with quotes), which do the same as if "" ; then ….Diabolo
O
15

See section 2.9.1 of the language spec. The last sentence of the first section reads:

If there is a command name, execution shall continue as described in Command Search and Execution . If there is no command name, but the command contained a command substitution, the command shall complete with the exit status of the last command substitution performed. Otherwise, the command shall complete with a zero exit status.

The $(true) is expanding to the empty string. The shell parses the empty string and finds that no command is given and follows the above rule.

Oscillograph answered 23/1, 2012 at 14:30 Comment(4)
Excellent! That's easy to miss (I did).Devil
Hey, I was right! Thanks for finding the pertinent section in the specRockandroll
Is there a guarantee in the standard that "if $(true) $(false)" will evaluate the false last?Oscillograph
@WilliamPursell: You asked, “Is there a guarantee in the standard that if $(true) $(false) will evaluate the false last?” The answer lies in the second sentence of Section 2.9.1: “When a given simple command is required to be executed …, the following expansions, assignments, and redirections shall all be performed from the beginning of the command text to the end” (emphasis added).Diabolo
L
6

There are potentially two exit codes to consider. First, here's another two experiments that should help:

# if $(echo true; false) ; then echo yes ; else echo no ; fi
yes

The inner command exits with failure because of the false. But that's irrelevant because the output of the command is non-empty and hence the output ("true") is executed instead and its exit code takes priority.

# if $(echo false; true) ; then echo yes ; else echo no ; fi
no

Again, the command line inside the $( ) is successful, but the output is no because the output ("false") takes priority.

The exit status of the commands inside the $( ) is relevant if and only if the output is empty or only whitespace. If the output is empty, there is no list of commands to execute and hence it appears that the shell will fall back on the exit status of the inner command.

Loriloria answered 23/1, 2012 at 0:30 Comment(4)
It's not so much a case of priority but the fact that bash only keeps one exit code and its value is the exit code of the last executed command. $(true) and $(false) are sub'd for null commands which does not count towards execution and thus the exit code of true and/or false remain.Rockandroll
"The exit status of the commands inside the $( ) is relevant if and only if the output is empty." That appears to be the case, but why? Can you point to any documentation from which you could infer that behavior?Devil
@KeithThompson, I don't know any such documentation. I was tempted to refer to this as an 'undocumented' feature :-) As @Rockandroll said, I think that it uses the exit status of the last command that was actually executed, regardless of in what 'context' that command was executed. i.e. Inside or outside $( ), what was the last command executed?Loriloria
“The exit status of the commands inside the $( ) is relevant if and only if the output is empty”: not exactly; it's enough that the output be all whitespace. See William Pursell's answer.Shad
R
5

What appears to be happening is that bash keeps the exit status of the last executed command

This would explain why $(true) and $(false) have different behavior in an if test. They both produce null commands which doesn't count as execution but they have different exit codes.

As soon as you use command substitution on a command that has output, $() attempts to execute that output as a command and the exit code of that attempt is now the latest one used for the if test

Rockandroll answered 23/1, 2012 at 0:29 Comment(0)
D
3

What should an if statement do when the condition is a command substitution where the command produces no output?

Output doesn't matter. What matters is the exit code:

   if list; then list; [ elif list; then list; ] ... [ else
   list; ] fi
          The if list is executed.  If its exit status is zero,
          the then list is executed.  Otherwise, each elif list
          is executed in turn, and if its exit status is zero,
          the corresponding then list is executed and the
          command completes.  Otherwise, the else list is
          executed, if present.  The exit status is the exit
          status of the last command executed, or zero if no
          condition tested true.

If you replace your $(echo true) and $(echo false) with something else you will probably see what is going on:

$ if $(echo false) ; then echo yes ; else echo no ; fi
no
$ if $(echo command-does-not-exist) ; then echo yes ; else echo no ; fi
command-does-not-exist: command not found
no
$ 

The $(..) runs the command and then if executes the results (in this case, just true or false or does-not-exist). An empty $() starts a subshell which successfully runs to completion and returns an exit code of 0:

$ if $() ; then echo yes ; else echo no ; fi
yes

Aaron raised some interesting points:

$ if $(echo true; false) ; then echo yes ; else echo no ; fi
yes
$ if $(echo 'true ; false') ; then echo yes ; else echo no ; fi
yes
$ if true ; false ; then echo yes ; else echo no ; fi
no

$ if $(echo false ; true) ; then echo yes ; else echo no ; fi
no
$ if $(echo 'false ; true') ; then echo yes ; else echo no ; fi
no
$ if false ; true ; then echo yes ; else echo no ; fi
yes
$ 

It appears that when executing a command constructed via the $() subshell, the first command's exit status is what matters.

Defile answered 23/1, 2012 at 0:13 Comment(8)
The exit code of the command inside the $( ) doesn't matter if the output is non-empty. if $(echo true; false) ; then echo yes ; else echo no ; fi gives yes. So the false is ignored because the outputed command ("true") is executed instead.Loriloria
@Aaron, that's a cool example. I had to play with it a bit further and now I'm more confused than when I started. :)Defile
It appears that when executing a command constructed via the $() subshell, the first command's exit status is what matters. If that were the case then the following should produce no: if $(false;echo 'true ; false') ; then echo yes ; else echo no ; fiRockandroll
@SiegeX: but the first command constructed via the $() subshell in your example is true. Don't forget that if executes the results of the output of the $() ... (If you can suggest better phrasing, please do. I went around on circles on that specific sentence for five minutes before settling on that.)Defile
"An empty $() starts a subshell which successfully runs to completion and returns an exit code of 0" -- That does appear to be the case; the question is why? Where is this behavior documented?Devil
@Defile So after looking at it a bit more, the reason why if $(echo 'true ; false'); then ... and if true ; false; then ... give different results is because command substitution only undergoes word splitting and glob expansion, nothing more. Therefore $(echo 'true ; false') expands to true \; false, i.e. true is called with two arguments: ; and false and not two separate commandsRockandroll
Perhaps this should be considered a special case of "if foo=$(cmd)", in which no variable assignments are being made.Oscillograph
@SiegeX’s explanation can be further demonstrated by if $(echo 'ls -l false ; true') ; then …, which outputs false: No such file, ;: No such file, and true: No such file (unless you have files by those names), and then no.  This shows that the ; is interpreted as an argument and not a command separator.  On the other hand, if $(echo 'eval false ; true') ; then … outputs yes, because the eval causes the shell to interpret the ; as a separator, and to execute false and then true.  (But don’t get in the habit of using eval; to the contrary, you should try to avoid it.)Diabolo
P
0

To add what a lot of people are probably trying to determine when looking for the reasons for this behavior:

If you want to strictly evaluate if a variable is boolean true, evaluating to false in all cases other than boolean true), use:

if [[ ${subject} = true ]]; then { ... } fi

If you want to strictly evaluate if a variable is boolean false, use:

if [[ ${subject} = false ]]; then { ... } fi
Plankton answered 16/3, 2023 at 14:9 Comment(0)

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