How to interrupt running Lua

script.lua

local x = 0
for i = 1, 10000000000 do
    local y = getNumber() -- C function
    x = x + y
end
print(x)

wrapper.cpp

// C++11
// Compile on Debian Wheezy:
// g++ -std=c++11 wrapper.cpp -o wrapper.exe \
//    -llua5.1 -I /usr/include/lua5.1/ -pthread

#include <cstdlib>
#include <cstdio>
#include <ctime>
#include <thread>

#include <lua.hpp>

bool stop_lua = false;

int getNumber(lua_State* L) {
    if (stop_lua) {
        return luaL_error(L, "Interrupted");
    }
    // return random number
    int value = std::rand();
    lua_pushinteger(L, value);
    return 1;
}

void luaRunner() {
    // this code is run in other thread
    lua_State* L = luaL_newstate();
    luaL_openlibs(L);
    // bind function getGlobal
    lua_pushcfunction(L, getNumber);
    lua_setglobal(L, "getNumber");
    // run script.lua
    int error = luaL_dofile(L, "script.lua");
}

int main() {
    std::srand(std::time(NULL));
    std::thread thread(luaRunner);
    printf("Lua works in other thread\n");
    while (true) {
        printf("Type 123 to interrupt Lua.\n");
        int input;
        scanf("%d", &input);
        if (input == 123) {
            stop_lua = true;
            printf("Lua stopped!.\n");
        }
    }
}

Nice idea!

But as a side-note for other readers signaling another thread this way is unsafe (it's not guaranteed that stop_lua will ever read the changed value in the "luaRunner" thread, for example, because CPU caches are never synchronized).
The simplest way would be using an std::atomic.