Bash: Compare exit codes of two functions

Question

Is there an elegant way to compare exit codes of two functions in bash? For example

b ()
{
    local r=$(rand -M 2 -s $(( $(date +%s) + $1 )) );
    echo "b$1=$r";
    return $r;
} # just random boolean    
b1 () { b 1; return $?; } # some boolean function
b2 () { b 2; return $?; } # another boolean function ( another seed )

I'd like to use something like this

if b1 == b2 ; then echo 'then'; else echo 'else'; fi

but stuck with this "not xor" implementation

if ! b1 && ! b2 || ( b1 && b2 ) ; then echo 'then'; else echo 'else'; fi

And speaking more generally, can one compare exit codes of two functions arithmetically and use that comparison in if statement?

Just to clarify, what do you expect this code b 1; return $?; to do? Since the result of running b 1 is not saved in a variable, what are you hoping to accomplish by running it? Separately, what do you expect the value of $? to signify? — John1024
– John1024, Commented May 23, 2016 at 22:16
I expect b 1; return $?; returns result of b 1, i.e. zero or one, because $? is a variable that stores exit code of last executed command. Yes, i do understand difference between stdout and function result; — Leonid
– Leonid, Commented May 23, 2016 at 22:38
Shell functions don't have return values; they have exit statuses, with well-defined semantics: 0 means success, any other non-zero value means failure. — chepner
– chepner, Commented May 23, 2016 at 23:03
Ahhh, i see why this question. It's just a sample. Actually b1 and b2 are some test functions like IsSomeAssertion and IsAnotherAssertion. And i want to know if both of them are false or true at the same time. Just now i end up with this solution: if (( $(b1)$? == $(b2)$? )) ; then echo 'then'; else echo 'else'; fi but it works only if b1 and b2 don't put anything to stdout. — Leonid
– Leonid, Commented May 23, 2016 at 23:08
@chepner yes, i meant return status. BTW, i never used term "return value" — Leonid
– Leonid, Commented May 23, 2016 at 23:14

John1024 · Accepted Answer · 2016-05-23 22:50:43Z

1

To compare exit codes of b1 and b2:

b1; code1=$?
b2; code2=$?
[ "$code1" -eq "$code2" ] && echo "exit codes are equal"

In shell, statements like b1 == b2 cannot stand alone. They need to be part of a test command. The test command is commonly written as [...]. Also, in shell, = (or, where supported, ==) are for string comparison. -eq is for numeric comparison.

answered May 23, 2016 at 22:50

John1024

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2 Comments

Leonid Over a year ago

Thank you for good explanation. By the way, as far as i understand, in (( )) construction == is for numeric comparison.

John1024 Over a year ago

Yes, inside ((...)), = is assignment and == is a test for equality. The construct ((...)) offers bash's full arithmetic capabilities. This, however, can lead to unpleasant surprises. Another difference is [...] is POSIX and therefore portable but ((...)) is not.

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