Calling C library function from Lua

Question

I'm trying to load the following C library in Lua:

#include <stdio.h>
#include <stdlib.h>
#include <lua.h>
#include <lualib.h>
#include <lauxlib.h>

void stackDump(lua_State* lua)
{
    int i, t;
    int top = lua_gettop(lua);

    printf("Stack dump: ");

    for (i = 1; i <= top; i++)
    {
        t = lua_type(lua, i);

        switch (t)
        {
            case LUA_TSTRING:
                printf("\"%s\"", lua_tostring(lua, i));
                break;

            case LUA_TBOOLEAN:
                printf(lua_toboolean(lua, i) ? "true" : "false");

            case LUA_TNUMBER:
                printf("%g", lua_tonumber(lua, i));
                break;

            default:
                printf("%s", lua_typename(lua, t));
                break;
        }

        if (i < top)
           printf(", ");
    }

    printf("\n");
}

int myPrint(lua_State* lua)
{
   stackDump(lua);

   const char* str = luaL_checkstring(lua, 1);

   printf("myPrint: %s\n", str);

   lua_newtable(lua);

   lua_pushinteger(lua, 1);
   lua_pushstring(lua, "myPrint");
   lua_settable(lua, -3);

   lua_pushinteger(lua, 2);
   lua_pushstring(lua, str);
   lua_settable(lua, -3);

   return 2;
}

static const struct luaL_Reg myScrewedLib[] = {
   {"myPrint", myPrint},
   {NULL, NULL}
};

int luaopen_myScrewedLib(lua_State* lua)
{
   luaL_newlib(lua, myScrewedLib);
   printf("SCREW IT\n");
   return 1;
}

I'm compiling this file with:

gcc -Wall -fpic -llua -shared -o myScrewedLib.so tut1.c

I see the following when running in the interpreter:

$ lua
Lua 5.2.2  Copyright (C) 1994-2013 Lua.org, PUC-Rio
> local m = require("myScrewedLib")
SCREW IT
> for k, v in pairs(m) do print(k, v) end
stdin:1: bad argument #1 to 'pairs' (table expected, got nil)
stack traceback:
    [C]: in function 'pairs'
    stdin:1: in main chunk
    [C]: in ?
>

I wouldn't expect m to be nil here. The book mentions the use of lua_pushcfunction() but only when it's a "quick-and-dirty way"[1] to run the function from the interpreter. Other examples of using the methods in this C file dont use lua_pushcfunction() as they imply that the luaL_Reg struct is how Lua knows what can be called[2].

[1] Programming in Lua 3rd Edition, p.274

[2] Programming in Lua 3rd Edition, p.279

Store your program in a file prior to running it `lua your_program.lua`. Or wrap whole program in `do ... end` if you want to use interpreter in interactive mode. — Egor Skriptunoff, Oct 01 '15 at 22:57
Yes, sorry I've just realised my big mistake; using the interpreter with a local variable. Should I submit an answer to my question? — Doddy, Oct 01 '15 at 22:58
you may simply close it if it's irrelevant, or if you think it's still useful for others, yes, you may self-answer it. — user3159253, Oct 01 '15 at 23:14

score 1 · Answer 1 · answered Oct 02 '15 at 07:56

I'm submitting the answer myself in case it is useful for those interested.

I was using the interactive interpreter to test my Lua. The issue with that is that each line you give to the interpreter is a separate chunk. So variables declared local immediately go out of scope after being entered.

The solution to this is to wrap your group of statements in a do .. end block, which ensures that the interpreter parses the whole block as one chunk.

Calling C library function from Lua

1 Answers1