lua调用函数来自一个具有函数名的字符串。

Is it possible in lua to execute a function from a string representing its name?
i.e: I have the string x = "foo", is it possible to do x() ?

在lua中是否可以从表示其名称的字符串中执行一个函数?我。e:我有字符串x = "foo"，可以做x()吗?

If yes what is the syntax ?

如果是，语法是什么?

5 个解决方案

#1

To call a function in the global namespace (as mentioned by @THC4k) is easily done, and does not require loadstring().

要在全局名称空间中调用函数(如@THC4k所述)很容易完成，而且不需要loadstring()。

x='foo'
_G[x]() -- calls foo from the global namespace

You would need to use loadstring() (or walk each table) if the function in another table, such as if x='math.sqrt'.

如果另一个表中的函数，比如x='math.sqrt'，则需要使用loadstring()(或遍历每个表)。

If loadstring() is used you would want to not only append parenthesis with ellipse (...) to allow for parameters, but also add return to the front.

如果使用loadstring()，您不仅需要使用椭圆(…)附加圆括号来允许参数，还需要添加返回到前面。

x='math.sqrt'
print(assert(loadstring('return '..x..'(...)'))(25)) --> 5

or walk the tables:

或走的表:

function findfunction(x)
  assert(type(x) == "string")
  local f=_G
  for v in x:gmatch("[^%.]+") do
    if type(f) ~= "table" then
       return nil, "looking for '"..v.."' expected table, not "..type(f)
    end
    f=f[v]
  end
  if type(f) == "function" then
    return f
  else
    return nil, "expected function, not "..type(f)
  end
end

x='math.sqrt'
print(assert(findfunction(x))(121)) -->11

#2

loadstring is not the answer here. For starters you would need a return in the string, and other details I won't go into.

loadstring不是这里的答案。对于初学者，你需要在字符串中返回，而其他的细节我不会进入。

THC4k has the right idea; if you have the function name in the variable x, then the call you want is

THC4k有正确的想法;如果在变量x中有函数名，那么需要的调用是。

_G[x](arg1, arg2, ...)

#3

I frequently put a bunch of functions in a table:

我经常把一些函数放在一个表中:

functions = {
       f1 = function(arg) print("function one: "..arg) end,
       f2 = function(arg) print("function two: "..arg..arg) end,
       ...,
       fn = function(arg) print("function N: argh") end,
}

Then you can use a string as an table index and run your function like this

然后，可以使用字符串作为表索引，并像这样运行函数。

print(functions["f1"]("blabla"))
print(functions["f2"]("blabla"))

This is the result:

这是由于:

function one: blabla
function two: blablablabla

I find this to be cleaner than using loadstring(). If you don't want to create a special function table you can use _G['foo'].

我发现这比使用loadstring()更干净。如果您不想创建一个特殊的函数表，可以使用_G['foo']。

#4

Names are not unique, there can be many functions names foo in different namespaces. But _G['foo'] is foo in the global namespace.

名称不是唯一的，在不同的名称空间中可以有很多函数名foo。但是_G['foo']在全局名称空间中是foo。

#5

It sounds like you want to do an 'eval', which is supported in Lua like so:

听起来你想要做一个“eval”，它在Lua中得到支持:

assert(loadstring(x))()

You'll probably want to concatenate the "()" onto x first, though.

不过，您可能想先将“()”连接到x上。

#1