Lua协同例程

Lua co-routines

本文关键字：协同例程 Lua      更新时间：2023-10-16

我正试图了解如何使用协同例程来"暂停"脚本，并等待一些处理完成后再恢复。

也许我以错误的方式看待协同例程。但我的尝试结构类似于这个答案中给出的例子。

loop.lua中的循环从未达到第二次迭代，因此从未达到C代码中退出运行循环所需的i == 4条件。如果我在loop.lua中没有屈服，则此代码按预期执行。

main.cpp 

#include <lua/lua.hpp>
bool running = true;
int lua_finish(lua_State *) {
    running = false;
    printf("lua_finish calledn");
    return 0;
}
int lua_sleep(lua_State *L) {
    printf("lua_sleep calledn");
    return lua_yield(L,0);
}
int main() {
    lua_State* L = lua_open();
    luaL_openlibs(L);
    lua_register(L, "sleep", lua_sleep);
    lua_register(L, "finish", lua_finish);
    luaL_dofile(L, "scripts/init.lua");
    lua_State* cL = lua_newthread(L);
    luaL_dofile(cL, "scripts/loop.lua");
    while (running) {
        int status;
        status = lua_resume(cL,0);
        if (status == LUA_YIELD) {
            printf("loop yieldingn");
        } else {
            running=false; // you can't try to resume if it didn't yield
            // catch any errors below
            if (status == LUA_ERRRUN && lua_isstring(cL, -1)) {
                printf("isstring: %sn", lua_tostring(cL, -1));
                lua_pop(cL, -1);
            }
        }
    }
    luaL_dofile(L, "scripts/end.lua");
    lua_close(L);
    return 0;
}

loop.lua 

print("loop.lua")
local i = 0
while true do
    print("lua_loop iteration")
    sleep()
    i = i + 1
    if i == 4 then
        break
    end
end
finish()

编辑:添加了一个赏金，希望能得到一些关于如何完成这个任务的帮助。

lua_resume返回代码2为LUA_ERRRUN。检查Lua堆栈顶部的字符串以查找错误消息。

一个类似的模式已经为我工作，虽然我确实使用coroutine.yield而不是lua_yield，我在C而不是c++。我不明白为什么你的解决办法行不通

在你的简历电话中，我不清楚你是否过度简化了一个例子，但我会在你的while循环中做以下改变:

int status;
status=lua_resume(cL,0);
if (status == LUA_YIELD) {
  printf("loop yieldingn");
}
else {
  running=false; // you can't try to resume if it didn't yield
  // catch any errors below
  if (status == LUA_ERRRUN && lua_isstring(cL, -1)) {
    printf("isstring: %sn", lua_tostring(cL, -1));
    lua_pop(cL, -1);
  }
}
编辑2:

要进行调试，请在运行简历之前添加以下内容。你有一个字符串被压入堆栈的某个地方:

int status;
// add this debugging code
if (lua_isstring(cL, -1)) {
  printf("string on stack: %sn", lua_tostring(cL, -1));
  exit(1);
}
status = lua_resume(cL,0);
编辑3:

哦，好悲伤，我不敢相信我没有看到这个，你不想运行luaL_dofile当你要屈服的时候，因为你不能直接屈服pcall据我所知，这是在dofile中发生的事情(5.2将传递它，但我认为你仍然需要lua_resume)。切换到这个:

luaL_loadfile(cL, "scripts/loop.lua");

上次我摆弄Lua协程时，我写了这样的代码

const char *program =
"function hello()n"
"  io.write("Hello world 1!")n"
"  io.write("Hello world 2!")n"
"  io.write("Hello world 3!")n"
"endn"
"function hate()n"
"  io.write("Hate world 1!")n"
"  io.write("Hate world 2!")n"
"  io.write("Hate world 3!")n"
"endn";
const char raw_program[] = 
"function hello()n"
"  io.write("Hello World!")n"
"endn"
"n"
"cos = {}n"
"n"
"for i = 0, 1000, 1 don"
"  cos[i] = coroutine.create(hello)n"
"endn"
"n"
"for i = 0, 1000, 1 don"
"  coroutine.resume(cos[i])n"
"end";
int _tmain(int argc, _TCHAR* argv[])
{
    lua_State *L = lua_open();
    lua_State *Lt[1000];
    global_State *g = G(L);
    printf("Lua memory usage after open: %dn", g->totalbytes);
    luaL_openlibs(L);
    printf("Lua memory usage after openlibs: %dn", g->totalbytes);
    lua_checkstack(L, 2048);
    printf("Lua memory usage after checkstack: %dn", g->totalbytes);
    //lua_load(L, my_lua_Reader, (void *)program, "code");
    luaL_loadbuffer(L, program, strlen(program), "line");
    printf("Lua memory usage after loadbuffer: %dn", g->totalbytes);
    int error = lua_pcall(L, 0, 0, 0);
    if (error) {
        fprintf(stderr, "%s", lua_tostring(L, -1));
        lua_pop(L, 1);
    }
    printf("Lua memory usage after pcall: %dn", g->totalbytes);
    for (int i = 0; i < 1000; i++) {
        Lt[i] = lua_newthread(L);
        lua_getglobal(Lt[i], i % 2 ? "hello" : "hate");
    }
    printf("Lua memory usage after creating 1000 threads: %dn", g->totalbytes);
    for (int i = 0; i < 1000; i++) {
        lua_resume(Lt[i], 0);
    }
    printf("Lua memory usage after running 1000 threads: %dn", g->totalbytes);
    lua_close(L);
    return 0;
}

似乎你不能加载文件作为协程?但是用function代替，它应该被选择到堆栈的顶部。

lua_getglobal(Lt[i], i % 2 ? "hello" : "hate");
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