Why does this shell script fail if I split it in two lines?

Question

I'm very comfortable using the shell, but I don't know much about shell scripting. Today I cam across a tiny shell script for fixing the node-webkit libudev.so.0 problem: https://github.com/rogerwang/node-webkit/wiki/The-solution-of-lacking-libudev.so.0

As you can see, solution one instructs to create a shell script with the following code in it:

LD_LIBRARY_PATH=/home/omi/nw:$LD_LIBRARY_PATH ./nw $*

My understanding of what is happening is the following:

1) Declare the variable LD_LIBRARY_PATH, and assign a value to it, in this case it is the path to the nw directory where we have previously created a symlink to libudev.so.1. Then, append the previous value of $LD_LIBRARY_PATH to the new value. Then execute the nw binary passing the arguments that were passed to the shell script.

This works well, as expected, but if I split the code in two lines like so:

LD_LIBRARY_PATH=/home/omi/nw:$LD_LIBRARY_PATH
./nw $*

It does not work. The shell script executes without any errors, but the node-webkit application doesn't run. Instead we get the error about libudev.so.0 missing. Why is that?

Also, it seems like the path you add (/home/omi/nw in this case) can be anything. I tried using /home/, /home/akjfd/, and other variations, and the script always worked as expected. Why is that? (I finally left it with the correct path pointing to where I created the symlink)

PS. A bit out of topic, I found out that creating a symlink to libudev.so.1 in the same directory (/lib64/ in Fedora 18 64bit) and calling it libudev.so.0 works as well without having to create the shell script, but I would not recommend it since you would be polluting the /lib64/ directory, and it may cause problems in your system.

Answer 1

Commands that are prefixed by a variable assignment run in a modified environment.

LD_LIBRARY_PATH=/home/omi/nw:$LD_LIBRARY_PATH ./nw $*

is (roughly) equivalent to

OLD_PATH="$LD_LIBRARY_PATH"
export LD_LIBRARY_PATH=/home/omi/nw:$LD_LIBRARY_PATH
./nw $*
LD_LIBRARY_PATH="$OLD_LIBRARY_PATH"

./nw runs in an environment where LD_LIBRARY_PATH is set to the given value, but nothing in the current environment is changed.

Your two-line command is equivalent to

LD_LIBRARY_PATH=/home/omi/nw:$LD_LIBRARY_PATH; ./nw $*

(note the semicolon). If LD_LIBRARY_PATH wasn't already marked for export to the environment, it is simply a local shell variable, and is not inherited by ./nw when it runs.

Answer 2

Here's your first script:

LD_LIBRARY_PATH=/home/omi/nw:$LD_LIBRARY_PATH ./nw $*

That command sets LD_LIBRARY_PATH in the environment of the child process (which runs nw).

Here's your second script:

LD_LIBRARY_PATH=/home/omi/nw:$LD_LIBRARY_PATH
./nw $*

That script sets the LD_LIBRARY_PATH shell variable. It doesn't modify the environment of any process, unless the shell is exporting LD_LIBRARY_PATH, which doesn't happen by default. You need to ask the shell to export LD_LIBRARY_PATH at least once to make that happen, using the export command. Example:

export LD_LIBRARY_PATH=/home/omi/nw:$LD_LIBRARY_PATH
./nw $*

You can also export separately from setting the variable, like this:

LD_LIBRARY_PATH=/home/omi/nw:$LD_LIBRARY_PATH
export LD_LIBRARY_PATH
./nw $*