c++ lua error on setting global variable

Question

I have the folowing code:

lua_State *lua;
lua = lua_open();
luaL_openlibs(lua);

std::string code = "print(gvar)\n"
    "function test()\n"
    "print(gvar)\n"
    "end\n";

if(!luaL_loadstring(lua, code.c_str())){
    if (lua_pcall(lua, 0, 0, 0)){
        const char* error = lua_tostring(lua, -1); 
        lua_pop(lua, 1);
    }
}

lua_pushstring(lua, "100");
lua_setglobal(lua, "gvar");
if (lua_pcall(lua, 0, 0, 0)){
    const char* error = lua_tostring(lua, -1); // returns "attempt to call a nil value"
    lua_pop(lua, 1);
}

lua_close(lua);

Calling functions and getting global variables works fine, but when i try to set global variable i get "attempt to call a nil value". And i cant understand why is that?

Caladan · Accepted Answer · 2013-08-07 11:27:24Z

2

if(!luaL_loadstring(lua, code.c_str())){
    if (lua_pcall(lua, 0, 0, 0)){
        const char* error = lua_tostring(lua, -1); 
        lua_pop(lua, 1);
    }
}

This code loads string into a anonymous function using luaL_loadstring(), puts it on the stack and then executes the function using lua_pcall(lua, 0, 0, 0).

lua_pushstring(lua, "100");
lua_setglobal(lua, "gvar");
if (lua_pcall(lua, 0, 0, 0)){
    const char* error = lua_tostring(lua, -1); // returns "attempt to call a nil value"
    lua_pop(lua, 1);
}

This piece of code pushes string onto the stack then sets global variable gvar. There should be nothing on the stack after call to lua_setglobal(). The var is already there.

Now after that you try to call a function which is at the top of the stack with lua_pcall, but the stack is empty - that's why you get attempt to call a nil value message.

answered Aug 7, 2013 at 11:27

Caladan

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2 Comments

Stals Over a year ago

oh, i see, thank you. I thought i needed to call lua_pcall so that variable will actually be added.

Caladan Over a year ago

Nope, you use lua_pcall() to call functions, all other things are just executed.

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