Bash function checks

Faki · 12-22-2021, 01:34 PM

I have a bash function that uses echo and sets a return value before exiting. I am using the following to set the variable var to the value sent to standard output.

Code:

var=$( myfunc --opt $arg )

I also use this inside if statements

Code:

var=$( myfunc --opt $arg )
if (( var == 1 )) && echo "something"


if myfunc --opt $arg 
then
  echo "do something"
fi

Both of these constructs use the standard output used by "echo".

For using the exit status, is it always the case that one uses "$?" ?

I would use in this way

Code:

$( myfunc --opt $arg )
if (( $? > 0 )) then
  echo "do something else"
fi

Here is myfunc

Code:

myfunc ()
{
 # do things here
 echo "$vb"
 return "$opst"
}

shruggy · 12-22-2021, 02:28 PM

Quote:

Originally Posted by Faki

For using the exit status, is it always the case that one uses "$?"

No. For me, this is rather an exception. I prefer AND (&&) and OR (||) lists, or using the command itself as part of conditional construct.

michaelk · 12-22-2021, 02:56 PM

By definition the function exit status is either the value from the last run command or as specified using the return keyword. The value is 0 for success or 1-255 for failure and is assigned to $?.
The returned value via echo command is not limited to 0-255 values.

Do something if myfunc exit status=0

Code:

if myfunc --opt $arg 
then
  echo "do something"
fi

In the case command 1 && command 2, command 2 will only run if command's 1 exit status=0.
In the case command 1 || command 2, command 2 will only run if command's 1 exit status not 0.

pan64 · 12-23-2021, 01:50 AM

Code:

# do not need $( command args )
command args
if (( $? > 0 )) then
....
fi
# and
if command args; then
....
fi

are identical
you can only return a single byte (from a function) = exit code of that function, return "string" is pointless, will not do what you want.