Luac don't work with one script

Question

I have embedded lua and I want precompile my script. For that, I call the main of luac (with argc the number of file is 1). My problem is on the function doargs of luac. I don't understand the use of the variable i. Because when I use one script. The result of i after the doargs function is 1. And in the main function we have argc -= i after. And so argc = 0 and I have a error "no file". Any idea ?

Answer 1

luac is meant to be a command line utility for compiling .lua files. This expected usage is the reason why you're getting an error.

When you start an executable the OS passes the name of the program as its first argument (argv[0]). The luac main function assumes it is being called by the OS, so it expects that there will always be at least one argument and its argv[0] will be the name of the executable.

For this reason doargs starts its for loop at 1 and will always ignore that first (0th) argument. It returns how many options it has processed, which is also the offset of the first filename in the argv array. The main function uses this to know where the list of files starts.

If you really want to use the main function to precompile your scripts, then supply an extra dummy argument at the beginning of your argument array and list your files after that. Preferably you should use luac from the command line and supply an output file where the precompiled script will be stored like this:

luac -o outputFile script.lua

Alternatively, take a look at chapter 8 of Programming in Lua (Compilation, Execution, and Errors) for a pure Lua solution, or the the luaL_dofile, luaL_dostring, lua_dump, and lua_load functions in the Reference Manual for a C API solution.