CODEWITHSUNDEEP
iosobjective-cxcode7.3swift3
Shanmugaraja G
Shanmugaraja G

Reputation: 2778

Swift - Check nil of an unwrapped optional variable

I am having a model class as follows, which is in a library written in Objective-C. I am consuming this class in my swift project. In swift it becomes property of type String!. Sometimes that property will be nil. So, I am testing the nil validation as follows:

Vendor.h

@interface Vendor: NSObject {

@property (nonatomic, strong) NSString *firstName;
@property (nonatomic, strong) NSString *lastName;
@property (nonatomic, strong) NSString *middleName;
}

In my Swift project, I am checking the nil validation for the middleName property as below:

if anObject.middleNam != nil { // Here, It throws runtime error: fatal error: Unexpectedly found nil while unwrapping an Optional value
}

It throws me an following runtime error:

fatal error: unexpectedly found nil while unwrapping an Optional value

If the objective-C properties exposed in swift as String? then I would have used the following:

if let middleName = anObject.middleName {

}

How would I check for the unwrapped optional variable.

Thanks in advance.

Upvotes: 7

Views: 2013

Answers (2)

Julien Quere
Julien Quere

Reputation: 2459

As Anil mentioned, the best solution would be to edit your objective-c code to add some _Nullable. But, as I understand, your Objective-C code is a library that you cannot edit. So, you have to deal with these String! that can be nil.

But you can simply use if let technique like this: 

if let firstName = vendor.firstName {
    print("Here is my firstName: \(firstName)")
} else {
    print("I have no firstName")
}
if let middleName = vendor.middleName {
    print("Here is my middleName: \(middleName)")
} else {
    print("I have no middleName")
}

if let lastName = vendor.lastName {
    print("Here is my name: \(lastName)")
} else {
    print("I have no lastName")
}

With this Vendor code, it returns the following result: 

@interface Vendor: NSObject
    @property (nonatomic, strong) NSString *firstName;
    @property (nonatomic, strong) NSString *lastName;
    @property (nonatomic, strong) NSString *middleName;
@end
@implementation Vendor
- (instancetype)init
{
    self = [super init];
    if (self) {
        self.firstName = @"Julien";
        self.middleName = nil;
        self.lastName = @"Quere";
    }

    return self;
}
@end

Result:

Here is my firstName: Julien 
I have no middleName 
Here is my name: Quere

Upvotes: 3

Anil Varghese
Anil Varghese

Reputation: 42977

If you want ObjectiveC property to be exposed in Swift as optional, mark it with _Nullable tag like 

@property (nonatomic, strong) NSString * _Nullable middleName;

Now middleName name would be optional of type String? instead of String! and you could conditionally unwrap it.

Read more about Nullability in ObjC

Upvotes: 6

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