如何使用gopherlua定义一个Lua函数，该函数有一个预定义的表作为param，Lua脚本可以在其中访问该函数中的表

基于@Koyaanisqatsi的回答，我发现了如何在围棋中让事情运转起来.

GO代码示例:

package main

import (
    "fmt"

    "github.com/yuin/gopher-lua"
)

type Person struct {
    Name       string
    GivenName  string
    Street     string
    PostalCode string
    City       string
}

func main() {
    p := &Person{
        Name:       "Mustermann",
        GivenName:  "Max",
        Street:     "Sackgasse 19",
        PostalCode: "36304",
        City:       "Alsfeld",
    }

    L := lua.NewState()
    defer L.Close()

    if err := L.DoFile("sample.lua"); err != nil {
        panic(err)
    }

    t := L.NewTable()
    t.RawSetString("name", lua.LString(p.Name))
    t.RawSetString("given_name", lua.LString(p.GivenName))
    t.RawSetString("street", lua.LString(p.Street))
    t.RawSetString("postal_code", lua.LString(p.PostalCode))
    t.RawSetString("city", lua.LString(p.City))

    if err := L.CallByParam(lua.P{
        Fn:      L.GetGlobal("call_me"),
        NRet:    1,
        Protect: true,
    }, t); err != nil {
        panic(err)
    }

    ret := L.Get(-1) // returned value
    L.Pop(1)         // remove received value

    fmt.Println("The result of the Lua function is:", ret)
}

Sample.lua文件:

function call_me(tbl)
    print(tbl.name)
    print(tbl.given_name)
    print(tbl.street)
    print(tbl.postal_code)
    print(tbl.city)

    return 0
end

结果:

Mustermann
Max
Sackgasse 19
36304
Alsfeld
The result of the Lua function is: 0