问题
I understand how optionals work, but this is throwing me for a loop. I have a variable called num and I want to increment it, so I did the following:
var num:Int! = 0
num++ //ERROR - Unary operator ++ cannot be applied to an operand of type Int!
But for some reason Swift won't let me increment a force unwrapped Int, even though it is supposed to be treated like a regular Int with the capability for nil behind the scenes. So I tried the following, and it worked:
var num:Int! = 0
num = num + 1 //NO ERROR
However, based on the error message it gave me, I tried the following to make the increment operator still work:
var num:Int! = 0
num!++ //NO ERROR
My question is why does the first bit of code break, when the second and third bits of code don't? Also, since num is an Int!, shouldn't I be able to treat it like a regular Int? Lastly, since an Int! is supposed to be treated like a regular Int, how am I able to unwrap it in the third example? Thanks.
回答1:
This error occurs with all inout parameters and should be considered a bug.
The usual way how inout parameters work is that their getter gets called once and their setter at least once. In this case the getter returns an Int!:
let num: Int! = 0
let num2 = num // is inferred to be of type Int!
so the signature of the getter/setter is not the same as the function/operator expects. But the compiler should implicitly unwrap the value if it gets assigned to an Int or passed like so:
var num3 = 0 // is of type Int
num3 = num // gets automatically unwrapped
// also with functions
func someFunc(i: Int) {}
someFunc(num) // gets automatically unwrapped
Sidenote:
Almost the same error occurs if you want to pass an Int to a function with an inout parameter of type Int!. But here it is obvious why this doesn't work (logically): The setter of an Int never takes a nil.
回答2:
You are using wrong type Int!, use Int instead
来源：https://stackoverflow.com/questions/31763467/swift-increment-int-not-working