Why -l option in lua interpreter acts strangely? - lua

Question

You can execute a sequence of chunks by giving them all as arguments
to the stand-alone interpreter, with the -l option. For instance, if
you have a file a with a single statement x=1 and another file b with
the statement print(x), the command line
prompt> lua -la -lb
will run the chunk in a, then the one in b, which will print the
expected 1.
The above is from the following link: https://www.lua.org/pil/1.1.html. Yet, when I was trying it out, I got a syntax error.
So, in file a.lua, I have only one line, which is a=1. Then in file b.lua, I have also only one line print("the value of a is:",a) . Then,
:~$ lua -i -la -lb
Lua 5.2.4 Copyright (C) 1994-2015 Lua.org, PUC-Rio
the value of a is: true
>
:~$
:~$ lua -la -lb
the value of a is: true
Lua 5.2.4 Copyright (C) 1994-2015 Lua.org, PUC-Rio
>
Why did it print out "the value of a is: true"? rather than "the value of a is: 1"?
Any comments are greatly appreciated.

Answer 1

The error is because you are use a = 1 in your file a.lua, unlike in the example where a.lua contains x = 1 and b.lua contains print(x).
Using a rather then x means your using the same a variable being written to when -la completes, changing it to true.
This happens because the option -l name is equal to name = require("name"). When require completes, on a file that returns no result to it, require will return true.
your command lua -la -lb would look something like this to lua:
a = require("a") --this returning true after it completes
b = require("b") --this printing the value of `a` which will always be `true`
Sources:
Similar question answered on Lua-Users: command-line -l option issue
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