Lua 检查参数类型
示例
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, "t使你没用 " .. type(tab) .." somewhere else!"
end
return tab.bar
end这有几个含义：
print(kungfoo(20)) --> prints 'nil, t使你没用 number somewhere else!' if kungfoo(20) then print "good" 其他打印 "bad" end --> prints bad foo = kungfoo(20) or "bar" --> sets foo to "bar"
现在，我们可以使用所需的任何参数来调用函数，并且不会使程序崩溃。
-- 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做些什么呢？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