C#访问lua获取一个table中的函数——映射成interface中的函数
using UnityEngine;
using System.Collections;
using System.Collections.Generic;
using XLua;
using System;
namespace Tutorial
{
public class CSCallLua : MonoBehaviour
{
LuaEnv luaenv = null;
//必须是加标签和public的才能导出
[CSharpCallLua]
public interface Person222
{
string name
{
get;set;
}
int age
{
get;set;
}
int add(int a, int b);
}
void Start()
{
luaenv = new LuaEnv();
luaenv = new LuaEnv();
TextAsset luaScript = Resources.Load<TextAsset>("hello");
luaenv.DoString(luaScript.text);
Person222 person = luaenv.Global.Get<Person222>("person");
int a = person.add(1, 2);
}
void OnDestroy()
{
luaenv.Dispose();
}
}
}
这里的add = function(self, a, b)必须有一个self作为第一个参数,否则会报错,原理未知。
若是写成:
person=
{
name ="xiaoming",
age = 12,
add = function (a, b) //这里没有self,则会报错
return a + b
end
}