How to compare result of function in if statement using idiomatic bash

Question

I have a function which returns one of three values, 0, 1 or 2

I want to check the return value of the function in an if statement.

How can I do this in idiomatic bash?

This works, but looks very wrong to me:

vercomp $prev $new
result=$?
if [[ result -eq 1 ]]; then
  echo "Greater than" 
elif [[ result -eq 2 ]]; then
  echo "Less than"
fi

How can I consolidate this code and use the function directly in the if statement instead of using the intermediate result variable and how can I access the return value without using $? ?

Everything I have tried has failed.

For example even this fails - why?

result=vercomp $prev $new
if [[ result -eq 1 ]]; then
  echo "Greater than" 
elif [[ result -eq 2 ]]; then
  echo "Less than"
fi

And this fails:

if [[ vercomp $prev $new -eq 1 ]]; then
  echo "Greater than" 
elif [[ vercomp $prev $new -eq 2 ]]; then
  echo "Less than"
fi

This fails:

if [[ $(vercomp $prev $new) -eq 1 ]]; then
  echo "Greater than" 
elif [[ $(vercomp $prev $new) -eq 2 ]]; then
  echo "Less than"
fi

What am I doing wrong? How can I improve my working code?

Inian · Accepted Answer · 2020-05-21T10:34:40.523

Bash functions return exit codes back to the shell which can be retrieved from the value of $? as in your first case. You need to read through Return value in a Bash function to understand the difference between exit code and return value from function.

As explained in detail in multiple answers including Function return values within BASH if statements you can directly use the function invocation in the if statement. A return code of 0 from the function would assert the if condition to true. So you can do something like

if vercomp "$prev" "$new"; then
    printf 'return code zero \n'
else
    case $? in
      1) printf 'return code one \n' ;;
      2) printf 'return code two \n' ;;
      *) printf "unknown return code $?\n" ;;
    esac
fi

As for the failure cases

This result=vercomp $prev $new is not even a valid function execution in bash but an incorrect variable assignment. To run function and store the output in a variable, you need to use Command Substitution $(..) You need to have written it as result=$(vercomp "$prev" "$new")
And if [[ vercomp $prev $new -eq 1 ]]; then again is an incorrect invocation of the function. Everything inside [[..]] is evaluated in a string context. So the test operator [[ just treats it as incorrect operation and would have thrown a parser error
[[ $(vercomp $prev $new) -eq 1 ]] is also incorrect. All commands in bash return an exit code to the shell the program is invoked from. The command substitution inherently returns the code to the shell via $? but the -eq operator checks for a string value returned from the invocation. Since the construct does not put any contents to stdout, there won't anything to match with 1 or 2. If you done soemthing like [[ $(vercomp "$prev" "$new"; printf "$?") -eq 1 ]] it would have worked, but again not a good practice.

Is there anyway to make it work without using the $? variable? — Nate Houk, May 21 '20 at 11:09
Yes but you can, using my first approach `if `, but that won't get you what the other return codes are. but what is your real problem you are trying to solve? — Inian, May 21 '20 at 11:11
Even with your answer ```if ``` you still need to use ```$?``` for any non-zero result. It seems strange to me that there is no way to check the return value without using ```$?```. But maybe that is just how Bash is, and it is my lack of understanding which makes it look confusing. — Nate Houk, May 21 '20 at 13:06

score 1 · Answer 2 · answered May 21 '20 at 10:11

1

With bash:

case $? in
  0) echo "one";;
  1) echo "two";;
  2) echo "three";;
esac

See: help case

answered May 21 '20 at 10:11

Cyrus

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How to compare result of function in if statement using idiomatic bash

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