Lua: compile a script with all dependencies

Question

Is there a way one can compile a script in which everything except standard Lua libraries is linked statically?

I am embedding a script in my C program, the functionality is split between modules, which are then loaded in main module. I would like to compile the main module into Lua bytecode, convert it to hex code, so it can easily be stored in my program. The problem is that main module still requires source code of other modules to be present either in LUA_PATH or in current directory. Is there a way how to override this behaviour? Sure, I could simply merge all files (automatically, during the compilation) into one lua source file, but since I am a beginning with Lua, I would like to know if there are some other solutions to this problem, and perhaps expand my horizons.

I was looking at luac and luajit but I could not make them to do what I need.

Any hints?

Answer 1

There are tools like srlua that may help. You may also check this presentation on luawrap and this discussion for ideas.

Answer 2

serialise lua_state

In C if you iterate through the loaded lua_state for functions and values you could create an as-if lua representation.

This could be serialised to lua form and then luac compiled. You would have to ignore C bound functions and userdata.

You would need to walk meta tables.

encapsulated form

By changing the loader, you could have a single resource which has each of lua files by name and treats load requests as seeks and reads in resource.