检查参数类型

某些函数仅适用于某种类型的参数：

function foo(tab)
    return tab.bar
end
--> returns nil if tab has no field bar, which is acceptable
--> returns 'attempt to index a number value' if tab is, for example, 3
--> which is unacceptable

function kungfoo(tab)
    if type(tab) ~= "table" then
        return nil, "take your useless " .. type(tab) .." somewhere else!"
    end

    return tab.bar
end

这有几个含义：

print(kungfoo(20)) --> prints 'nil, take your useless number somewhere else!'

if kungfoo(20) then print "good" else print "bad" end --> prints bad

foo = kungfoo(20) or "bar" --> sets foo to "bar"

现在我们可以使用我们想要的任何参数调用该函数，并且它不会使程序崩溃。

-- if we actually WANT to abort execution on error, we can still do
result = assert(kungfoo({bar=20})) --> this will return 20
result = assert(kungfoo(20)) --> this will throw an error

那么，如果我们有一个函数可以对特定类的实例执行某些操作呢？这很困难，因为类和对象通常是表，所以 type 函数将返回'table'。

local Class = {data="important"}
local meta = {__index=Class}

function Class.new()
    return setmetatable({}, meta)
end
-- this is just a very basic implementation of an object class in lua

object = Class.new()
fake = {}

print(type(object)), print(type(fake)) --> prints 'table' twice

解决方案：比较元表

-- continuation of previous code snippet
Class.is_instance(tab)
    return getmetatable(tab) == meta
end

Class.is_instance(object) --> returns true
Class.is_instance(fake) --> returns false
Class.is_instance(Class) --> returns false
Class.is_instance("a string") --> returns false, doesn't crash the program
Class.is_instance(nil) --> also returns false, doesn't crash either