有没有办法确定Lua函数的签名? [英] Is there a way to determine the signature of a Lua function?
问题描述
最近, Lee Baldwin 展示了如何编写通用的,可变参数memoize函数。我认为在只需要一个参数的情况下返回一个更简单的函数会更好。这是我的假试图:
本地函数memoize(f)
本地缓存= {}
if select('#',...)== 1然后
返回函数(x)
如果缓存[x]然后
返回缓存[x]
else
local y = f(x)
cache [x] = y
return y
end
end
else
return function (...)
local al = varg_tostring(...)
if cache [al] then b $ b return cache [al]
else
local y = f (...)
cache [al] = y
return y
end
end
end
end
显然,在这种情况下, select('#',...)
失败,无论如何我都不会真正做我想做的事。有什么办法可以告诉 memoize 有多少个参数 f 需要?
如果您确实知道,否是一个很好的答案。使用两个单独的 memoize 函数并不重要。
调试信息并从源代码中确定它,但基本上它是一个不,对不起。Recently, Lee Baldwin showed how to write a generic, variable argument memoize function. I thought it would be better to return a simpler function where only one parameter is required. Here is my total bogus attempt:
local function memoize(f)
local cache = {}
if select('#', ...) == 1 then
return function (x)
if cache[x] then
return cache[x]
else
local y = f(x)
cache[x] = y
return y
end
end
else
return function (...)
local al = varg_tostring(...)
if cache[al] then
return cache[al]
else
local y = f(...)
cache[al] = y
return y
end
end
end
end
Obviously, select('#', ...)
fails in this context and wouldn't really do what I want anyway. Is there any way to tell inside memoize how many arguments f expects?
"No" is a fine answer if you know for sure. It's not a big deal to use two separate memoize functions.
I guess you could go into the debug info and determine this from the source-code, but basically it's a "no", sorry.
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