CODEWITHSUNDEEP
swiftscopedeclare
RichardYY
RichardYY

Reputation: 93

Why outside of block swift can't see value assigned to a uninitialized variable in the block?

What is the mechanism of declaring w/o value in Swift5 ? Does the first assign become the real declaration ?

And, should we avoid to declare without value in Swift? 

var v:String;
if true {
    v = "Hello"
    print(v) // print "Hello" when without the print below
}
print(v) // variable 'v' used before being initialized

var v:String="";
if true {
    v = "Hello"
    print(v) // print "Hello"
}
print(v) // print "Hello"

Upvotes: 1

Views: 126

Answers (2)

matt
matt

Reputation: 535247

Well, the message is not very helpful, and that's the problem. This pattern (which I call computed initialization) is perfectly legal and useful and — amazingly — you can even use let instead of var. But you must initialize the uninitialized variable by all possible paths before you use it. So you have:

var v:String
if true {
    v = "Hello"
}
print(v) // error

But hold my beer and watch this:

var v:String
if true {
    v = "Hello"
} else {
    v = "Goodbye"
}
print(v) // fine!

Or even:

let v:String
if true {
    v = "Hello"
} else {
    v = "Goodbye"
}
print(v) // fine!

Amazing, eh?

Now, you might say: OK, but true will always be true so it's silly to make me fulfill the "all paths" rule. Too bad! The compiler insists anyway, and then lets you off later with a warning that the else won't be executed. But a warning lets you compile; an error doesn't. The truth is that your example is very artificial. But this is a real-life possibility:

let v:String
if self.someBoolProperty {
    v = "Hello"
} else {
    v = "Goodbye"
}
print(v) // fine!

Not only is this sort of thing legal, it is actually the pattern that Apple recommends under certain slightly tricky circumstances. For instance, it is used in Apple's own example code showing how to use the Swift 5 Result struct:

let result: Result<Int, EntropyError>
if count < AsyncRandomGenerator.entropyLimit {
    // Produce numbers until reaching the entropy limit.
    result = .success(Int.random(in: 1...100))
} else {
    // Supply a failure reason when the caller hits the limit.
    result = .failure(.entropyDepleted)
}

Upvotes: 4

Travis Delly
Travis Delly

Reputation: 1374

So this is because swift compiles your code and notices that your value var v:String; is undeclared before being used which makes it an "optional" value. Even though you are assigning it within the if statement, if you were to get rid of the true value it is possible that the if statement would never run therefore no value will ever be stored in v, thus it would be used before "assigned".

So to answer your question if you want your value to be an optional and possible empty value declare v as the following var v:String? if you would like it to be a non-optional value with a value always stored within v you should declare it as the following var v = "". Swift will interpret this declaration as a String.

To answer your second question, defining without values in swift is 100% allowed, it really just depends on how you want to handle your code. I use optional values in my code all the time, but I don't necessarily like optionals so i usually like to declare my values such as var v = "" that way if the value is empty my UI or whatever else i'm manipulating won't break. But if i need to ensure a value is present i will have to make my value optional so i can use an if statement to check whether it's a valid value or not.

Shorter version of what I'm trying to say is, you are receiving the compiler warning because you are declaring v as a non-optional value rather than an optional value.

Upvotes: 0

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