bash: assignment AND condition test in nested statement?

Question

What I'm looking for is a compact line of code that uses the value of an expression twice in one nested statement

• for an assignment to a variable and

• a condition test of the variable's new value against another value

The questions are:

1. Is this compact form generally possible in bash?

2. If it is, do I have to fear negative side effects (unexpected results)?

You can certainly make the assignment before:

$ local bus_connection="$(lsblk -dn -o TRAN /dev/sdd)"
$ test "$bus_connection" = "usb" || { echo "not usb"; exit 1 }
$ echo "$bus_connection"
usb

--> Works!

If variable 'bus_connection' is only needed in case of success you also can do:

$ test "$(lsblk -dn -o TRAN /dev/$DEVICE)" = "usb" \
    && local bus_connection="usb" \
    || { echo "not usb"; exit 1 }
$ echo "$bus_connection"
usb

--> Also works! (Don't use 'local' on the command line!)

In C(++) and some other compiled high level languages you can do it more compact (let's suppose a c function 'lsblk'):

if( ( bus_connection = lsblk( ... ) ) == "usb" ) ; else {printf("not usb"); exit 1;}

In bash I tried:

$ test $(bus_connection="$(lsblk -dn -o TRAN /dev/sdd)") = "usb" || echo "not usb" 
bash: test: =: unary operator exspected.
not usb

--> Error message and wrong result!

I tested with an integer case and found:

$ sum=0
$ result=12
$ test $((sum=$sum+$result)) -gt 20 && echo "true"
$ test $((sum=$sum+$result)) -gt 20 && echo "true"
true

--> Works! Negative side effects?

Analog:

$ test $((bus_connection="$(lsblk -dn -o TRAN /dev/sdd)")) = "usb" || echo "not usb"
not usb

--> No error message but wrong result!

$ echo "$bus_connection"
0

--> Ah, obviously bash treats variable 'bus_connection' as of type integer now.

Any suggestions for strings?

Answer 1

Is this compact form generally possible in bash

No, it is not, at least not without a custom code to handle it. Command substitution is run in a subshell - changes will not be visible in parent shell. Arithmetic expansion does only arithmetic operations - all strings are converted to numbers (as if by atoi()/strtol()), if no digits are scanned they are converted to 0.

The usual way is just to assign the variable with a && followed by test - that allows to check the exit status of the command and grab the result.

if ! {
     bus_connection="$(lsblk -dn -o TRAN /dev/sdd)" &&
     test "$bus_connection" = "usb"
}; then
   echo "not usb"
   exit 1
fi

or just ignore the exit status of lsblk:

if
   bus_connection="$(lsblk -dn -o TRAN /dev/sdd)"
   ! test "$bus_connection" = "usb"
then
   echo "not usb"
   exit 1
fi

Still you could write your own function that uses bash namereference to assign the first argument to the second and return with tests exit status:

test_and_assign() {
   declare -n _test_and_assign_ref=$1
   shift
   _test_and_assign_ref="$1"
   test "$@"
}

$ test_and_assign bus_connection "$(echo 123)" = "usb"
$ echo "$? $bus_connection"
1 123
$ test_and_assign bus_connection "$(echo usb)" = "usb"
$ echo "$? $bus_connection"
0 usb

Note that here, as in your first code snippet with local bus_connection=$(...), the exit status of the command substitution is ignored.