Lua 5.2版：'；尝试调用nil值'；来自lua_pcall 我在获取Lua 5.2函数时，遇到了问题，从C++调用。

这是Lua块（名为test.Lua）：

这就是C++：

int iErr = 0;

//Create a lua state
lua_State *lua = luaL_newstate();

// Load io library
luaopen_io (lua);

//load the chunk we want to execute (test.lua)
iErr = luaL_loadfile(lua, "test.lua");
if (iErr == 0) {
    printf("successfully loaded test.lua\n");

    // Push the function name onto the stack
    lua_getglobal(lua, "testFunction");
    printf("called lua_getglobal. lua stack height is now %d\n", lua_gettop(lua));

    //Call our function
    iErr = lua_pcall(lua, 0, 0, 0);
    if (iErr != 0) {
        printf("Error code %i attempting to call function: '%s'\n", iErr, lua_tostring(lua, -1));
    }

} else {
    printf("Error loading test.lua. Error code: %s\n", lua_tostring(lua, -1));        
}
lua_close (lua);

当我跟踪时，我看到它很好地加载了test.lua脚本（没有返回错误），然后在使用函数名调用lua_getglobal之后，它显示堆栈高度为3

但是，它在lua_pcall失败，错误代码为2:“尝试调用nil值”

我已经阅读了大量Lua5.2代码的示例，似乎看不出哪里出了问题。 这看起来肯定会起作用（根据我所读到的）

我检查了拼写和区分大小写的能力，结果都吻合

---

我误解了什么吗？

luaL\u loadfile

只加载文件，而不运行它。请尝试

luaL\u dofile

您仍然会得到一个错误，因为

print

是在基本库中定义的，而不是在io库中定义的。因此，请改为调用

luaopen\u base

。

您需要在

lua\u getglobal（）之前调用“
启动lua\u pacll（）
”。请参阅。整个代码应该是这样的：

int iErr = 0;

//Create a lua state
lua_State *lua = luaL_newstate();

// Load base library
luaopen_base (lua);

//load the chunk we want to execute (test.lua)
iErr = luaL_loadfile(lua, "test.lua");
if (iErr == 0) {
    printf("successfully loaded test.lua\n");

    //Call priming lua_pcall
    iErr = lua_pcall(lua, 0, 0, 0);
    if (iErr != 0) {
        printf("Error code %i attempting to call function: '%s'\n", iErr, lua_tostring(lua, -1));
    }

    // Push the function name onto the stack
    lua_getglobal(lua, "testFunction");
    printf("called lua_getglobal. lua stack height is now %d\n", lua_gettop(lua));

    //Call our function
    iErr = lua_pcall(lua, 0, 0, 0);
    if (iErr != 0) {
        printf("Error code %i attempting to call function: '%s'\n", iErr, lua_tostring(lua, -1));
    }

} else {
    printf("Error loading test.lua. Error code: %s\n", lua_tostring(lua, -1));        
}
lua_close (lua);

打印功能未加载<代码> LalalOpenOpenLBS 应该加载它。谢谢，我现在已经工作了，已经按照我的建议修改了C++。我现在了解到，要注册块中的函数，它必须是“运行”的，而不仅仅是Lua术语中的“加载”。谢谢你的回答@user2795503如果答案解决了问题，不要忘记“接受”答案。
int iErr = 0;

//Create a lua state
lua_State *lua = luaL_newstate();

// Load base library
luaopen_base (lua);

//load the chunk we want to execute (test.lua)
iErr = luaL_loadfile(lua, "test.lua");
if (iErr == 0) {
    printf("successfully loaded test.lua\n");

    //Call priming lua_pcall
    iErr = lua_pcall(lua, 0, 0, 0);
    if (iErr != 0) {
        printf("Error code %i attempting to call function: '%s'\n", iErr, lua_tostring(lua, -1));
    }

    // Push the function name onto the stack
    lua_getglobal(lua, "testFunction");
    printf("called lua_getglobal. lua stack height is now %d\n", lua_gettop(lua));

    //Call our function
    iErr = lua_pcall(lua, 0, 0, 0);
    if (iErr != 0) {
        printf("Error code %i attempting to call function: '%s'\n", iErr, lua_tostring(lua, -1));
    }

} else {
    printf("Error loading test.lua. Error code: %s\n", lua_tostring(lua, -1));        
}
lua_close (lua);