CODEWITHSUNDEEP
bash
Leonid
Leonid

Reputation: 118

Bash: Compare exit codes of two functions

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?

Upvotes: 1

Views: 1756

Answers (1)

John1024
John1024

Reputation: 113834

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.

Upvotes: 1

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