获得一个简单的Haskell HsLua示例

Question

Haskell Wiki上的HsLua示例已被破坏(dostring和dofile未定义).看起来自从编写示例以来API已经发生了变化.

但是,我一直在尝试修改示例以匹配当前的API,但没有取得多大成功.这里有一个程序,应该真正的工作!

main = do
    l <- Lua.newstate
    Lua.openlibs l
    Lua.loadfile l "configfile.lua"
    [name,pwd] <- Lua.callfunc l "getuserpwd" "mail.google.com"
    putStrLn name
    Lua.close l

但是,这甚至没有编译,给我这个奇怪的错误信息 -

No instance for (Lua.StackValue [String])
  arising from a use of `Lua.callfunc'
Possible fix:
  add an instance declaration for (Lua.StackValue [String])
In a stmt of a 'do' expression:
    [name, pwd] <- Lua.callfunc l "getuserpwd" "mail.google.com"
In the expression:
  do { l <- Lua.newstate;
       Lua.openlibs l;
       Lua.loadfile l "configfile.lua";
       [name, pwd] <- Lua.callfunc l "getuserpwd" "mail.google.com";
       .... }
In an equation for `main':
    main
      = do { l <- Lua.newstate;
             Lua.openlibs l;
             Lua.loadfile l "configfile.lua";
             .... }

虽然文件的内容configfile.lua可能无关紧要(因为haskell代码甚至没有编译),但它如下(与维基页面相同) -

function getuserpwd (site)
  local cookies = { ["www.ibm.com"] = {"joe", "secret"}
                  , ["www.sun.com"] = {"hoe", "another secret"}
                  }
  if cookies[site] then
    return cookies[site]
  elseif site:match("[.]google[.]com$") then
    return {"boss", "boss"}
  else
    return { os.getenv("USER") or "God"
           , os.getenv("PWD")  or "dontdisturb" }
  end
end

有人可以给我一个Haskell-> Lua和Lua-> Haskell调用的工作示例吗？

编辑

好吧,我将返回值的类型更改为String(来自早期的String数组),该程序确实编译了!但是它现在在运行时失败了.这是修改后的程序 -

main = do
    l <- Lua.newstate
    Lua.openlibs l
    Lua.loadfile l "configfile.lua"
    Lua.callfunc l "getuserpwd" "mail.google.com" >>= putStrLn
    Lua.close l

这是configfile.lua -

function getuserpwd (site)
  return "boss"
end

并且运行时错误消息如下 -

**** Exception: user error (attempt to call a nil value)

Answer 1

您必须在调用任何函数之前执行加载的Lua块:

main = do
    l <- Lua.newstate
    Lua.openlibs l
    Lua.loadfile l "configfile.lua"
    Lua.call l 0 0
    Lua.callfunc l "getuserpwd" "mail.google.com" >>= putStrLn
    Lua.close l

方法dofile是loadfile和call的包装器,我不知道删除它的原因.

编辑. 此代码调用返回表并迭代它的函数.它基于这个遍历示例.我不知道如何通过callfunc来做到这一点.

import qualified Scripting.Lua as Lua
import Control.Monad.Loops
import Data.Maybe

print_table l = do
    Lua.pushnil l
    whileM_ (Lua.next l 1) (Lua.tostring l (-1) >>= putStrLn >> Lua.pop l 1)

main = do
  l <- Lua.newstate
  Lua.openlibs l
  Lua.loadfile l "configfile.lua"
  Lua.call l 0 0
  Lua.getglobal l "getuserpwd"
  Lua.pushstring l "mail.google.com"
  Lua.call l 1 (-1) -- calls function without Haskell extensions
  print_table l
  Lua.close l

事实证明,HsLua实现是一个非常简单的包装器,并且Lua表没有合适的Haskell绑定.