Question

我有一个简单的Lua函数，我从一个修改另一个Lua文件中特定行的在线源中获取并修改了该函数。该函数似乎试图在我的控制台上弹出PANIC: unprotected error in call to Lua API (attempt to call a nil value)时调用nil值。此函数由另一个C ++函数调用。我对Lua几乎没有任何经验，也不知道发生了什么。

剧本：

function ModLine(value, line)
    print("Modding...")
    local file = io.open("Image/DM2240_HighScore.lua", "r")
    sContents = file:read() -- capture file in a string
    file:close()
    tContents = textutils.unserialize(sContents) -- convert string to table

    --Print a specific line
    print(tContents[line])

    --Modify a specific line
    table.remove(tContents, line) -- will remove line so we can insert the new line 
    table.insert(tContents, line, value) -- inserts value on line in the table.

    --Convert table to string and save the file
    sContents = textutils.serialize(tContents)
    file = io.open("Image/DM2240_HighScore.lua", "w")
    file:write(sContents)
    file:close()
end

提前致谢！

编辑：我检查了一些函数，发现问题出在这个函数中，

void LuaInterface::ModLine(const int value, const int line) {
    cout << "Modded" << endl;
    lua_getglobal(theLuaState, "ModLine");
    lua_pushinteger(theLuaState, value);
    lua_pushinteger(theLuaState, line);
    lua_call(theLuaState, 2, 0);
}

Lua脚本试图调用nil

0 个答案: