I'm using Xcode 6 Beta 4. I have this weird situation where I cannot figure out how to appropriately test for optionals.
If I have an optional xyz, is the correct way to test:
if (xyz) // Do something
or
if (xyz != nil) // Do something
The documents say to do it the first way, but I've found that sometimes, the second way is required, and doesn't generate a compiler error, but other times, the second way generates a compiler error.
My specific example is using the GData XML parser bridged to swift:
let xml = GDataXMLDocument(
XMLString: responseBody,
options: 0,
error: &xmlError);
if (xmlError != nil)
Here, if I just did:
if xmlError
it would always return true. However, if I do:
if (xmlError != nil)
then it works (as how it works in Objective-C).
Is there something with the GData XML and the way it treats optionals that I am missing?
Although you must still either explicitly compare an optional with nil
or use optional binding to additionally extract its value (i.e. optionals are not implicitly converted into Boolean values), it's worth noting that Swift 2 has added the guard
statement to help avoid the pyramid of doom when working with multiple optional values.
In other words, your options now include explicitly checking for nil
:
if xyz != nil {
// Do something with xyz
}
Optional binding:
if let xyz = xyz {
// Do something with xyz
// (Note that we can reuse the same variable name)
}
And guard
statements:
guard let xyz = xyz else {
// Handle failure and then exit this code block
// e.g. by calling return, break, continue, or throw
return
}
// Do something with xyz, which is now guaranteed to be non-nil
Note how ordinary optional binding can lead to greater indentation when there is more than one optional value:
if let abc = abc {
if let xyz = xyz {
// Do something with abc and xyz
}
}
You can avoid this nesting with guard
statements:
guard let abc = abc else {
// Handle failure and then exit this code block
return
}
guard let xyz = xyz else {
// Handle failure and then exit this code block
return
}
// Do something with abc and xyz
One option that hasn't specifically been covered is using Swift's ignored value syntax:
if let _ = xyz {
// something that should only happen if xyz is not nil
}
I like this since checking for nil
feels out of place in a modern language like Swift. I think the reason it feels out of place is that nil
is basically a sentinel value. We've done away with sentinels pretty much everywhere else in modern programming so nil
feels like it should go too.
From swift programming guide
If Statements and Forced Unwrapping
You can use an if statement to find out whether an optional contains a value. If an optional does have a value, it evaluates to true; if it has no value at all, it evaluates to false.
So the best way to do this is
// swift > 3
if xyz != nil {}
and if you are using the xyz
in if statement.Than you can unwrap xyz
in if statement in constant variable .So you do not need to unwrap every place in if statement where xyz
is used.
if let yourConstant = xyz{
//use youtConstant you do not need to unwrap `xyz`
}
This convention is suggested by apple
and it will be followed by devlopers.
Another approach besides using if
or guard
statements to do the optional binding is to extend Optional
with:
extension Optional {
func ifValue(_ valueHandler: (Wrapped) -> Void) {
switch self {
case .some(let wrapped): valueHandler(wrapped)
default: break
}
}
}
ifValue
receives a closure and calls it with the value as an argument when the optional is not nil. It is used this way:
var helloString: String? = "Hello, World!"
helloString.ifValue {
print($0) // prints "Hello, World!"
}
helloString = nil
helloString.ifValue {
print($0) // This code never runs
}
You should probably use an if
or guard
however as those are the most conventional (thus familiar) approaches used by Swift programmers.
One of the most direct ways to use optionals is the following:
Assuming xyz
is of optional type, like Int?
for example.
if let possXYZ = xyz {
// do something with possXYZ (the unwrapped value of xyz)
} else {
// do something now that we know xyz is .None
}
This way you can both test if xyz
contains a value and if so, immediately work with that value.
With regards to your compiler error, the type UInt8
is not optional (note no '?') and therefore cannot be converted to nil
. Make sure the variable you're working with is an optional before you treat it like one.
Now you can do in swift the following thing which allows you to regain a little bit of the objective-c if nil else
if textfieldDate.text?.isEmpty ?? true {
}
If you have conditional and would like to unwrap and compare, how about taking advantage of the short-circuit evaluation of compound boolean expression as in
if xyz != nil && xyz! == "some non-nil value" {
}
Granted, this is not as readable as some of the other suggested posts, but gets the job done and somewhat succinct than the other suggested solutions.
Swift 3.0, 4.0
There are mainly two ways of checking optional for nil. Here are examples with comparison between them
1. if let
if let
is the most basic way to check optional for nil. Other conditions can be appended to this nil check, separated by comma. The variable must not be nil to move for the next condition. If only nil check is required, remove extra conditions in the following code.
Other than that, if x
is not nil, the if closure will be executed and x_val
will be available inside. Otherwise the else closure is triggered.
if let x_val = x, x_val > 5 {
//x_val available on this scope
} else {
}
2. guard let
guard let
can do similar things. It's main purpose is to make it logically more reasonable. It's like saying Make sure the variable is not nil, otherwise stop the function. guard let
can also do extra condition checking as if let
.
The differences are that the unwrapped value will be available on same scope as guard let
, as shown in the comment below. This also leads to the point that in else closure, the program has to exit the current scope, by return
, break
, etc.
guard let x_val = x, x_val > 5 else {
return
}
//x_val available on this scope
Also you can use Nil-Coalescing Operator
The
nil-coalescing operator
(a ?? b
) unwraps an optionala
if it containsa
value, or returnsa
default valueb
ifa
isnil
. The expression a is always of an optional type. The expressionb
must match the type that is stored insidea
.
let value = optionalValue ?? defaultValue
If optionalValue
is nil
, it automatically assigns value to defaultValue
To add to the other answers, instead of assigning to a differently named variable inside of an if
condition:
var a: Int? = 5
if let b = a {
// do something
}
you can reuse the same variable name like this:
var a: Int? = 5
if let a = a {
// do something
}
This might help you avoid running out of creative variable names...
This takes advantage of variable shadowing that is supported in Swift.
var xyz : NSDictionary?
// case 1:
xyz = ["1":"one"]
// case 2: (empty dictionary)
xyz = NSDictionary()
// case 3: do nothing
if xyz { NSLog("xyz is not nil.") }
else { NSLog("xyz is nil.") }
This test worked as expected in all cases.
BTW, you do not need the brackets ()
.
Instead of if
, ternary operator might come handy when you want to get a value based on whether something is nil:
func f(x: String?) -> String {
return x == nil ? "empty" : "non-empty"
}
Swift 5 Protocol Extension
Here is an approach using protocol extension so that you can easily inline an optional nil check:
import Foundation
public extension Optional {
var isNil: Bool {
guard case Optional.none = self else {
return false
}
return true
}
var isSome: Bool {
return !self.isNil
}
}
Usage
var myValue: String?
if myValue.isNil {
// do something
}
if myValue.isSome {
// do something
}
Source: Stackoverflow.com