BASH: Passing result of boolean expression to a function

[demonGeek]

I'd like to evaluate a boolean expression and pass the result as a parameter to a function. Here's the code I have so far:


dayOfMonth=$( (date +%-d) )
[ $dayOfMonth = 10 ]
MyFunction "$?"The code above works but I'd really like to combine the boolean expression and the function call into a single line so that it looks something like this (which doesn't work):


MyFunction "[ $dayOfMonth = 10 ]"Any help would be appreciated.

- Adam

[croto]

How about changing the double quotes by backticks?


MyFunction `[ $dayOfMonth = 10 ]`


EDIT: My suggestion does not work, you need the exit code, not stdout.

[slakkie]

I don't think that is possible.

You could do it ala

bla=$( [ 0 -eq 0 ] ; echo $?)
your_func $bla

which would become:

your_func $([ 0 -eq 0 ] ; echo $?)

but that's the same as doing what you are currently doing. I would keep your current version.

[jpkotta]

How about changing the double quotes by backticks?


MyFunction `[ $dayOfMonth = 10 ]`


EDIT: My suggestion does not work, you need the exit code, not stdout.

Return status of the last command is $?. Zero means OK/true/no error.



false ; echo $?
[ 0 = 0 ] ; echo $?

[demonGeek]

Thanks for the help everyone.

Slakkie: Your suggestion worked out for me - it's ugly but it does the job:



dayOfMonth=$( (date +%-d) )
MyFunction "$([ $dayOfMonth = 10 ];echo $?)"