Assigning propertyWrapper

Hi, I am learning propertyWrappers in Swift last days and I found something strange on the internet. In this article https://nshipster.com/propertywrapper/ import Foundation @propertyWrapper struct CaseInsensitive<Value: StringProtocol> { var wrappedValue: Value } extension CaseInsensitive: Comparable { private func compare(_ other: CaseInsensitive) -> ComparisonResult { wrappedValue.caseInsensitiveCompare(other.wrappedValue) } static func == (lhs: CaseInsensitive, rhs: CaseInsensitive) -> Bool { lhs.compare(rhs) == .orderedSame } static func < (lhs: CaseInsensitive, rhs: CaseInsensitive) -> Bool { lhs.compare(rhs) == .orderedAscending } static func > (lhs: CaseInsensitive, rhs: CaseInsensitive) -> Bool { lhs.compare(rhs) == .orderedDescending } } struct Account: Equatable { @CaseInsensitive var name: String init(name: String) { $name = CaseInsensitive(wrappedValue: name) } } There is assigning propertyWrapper to the name inside the init and it looks strange for me why author is trying to assign to the projectedValue ($) instead of use _name to talk with propertyWrapped instance... Am I right that there should be: init(name: String) { _name = CaseInsensitive(wrappedValue: name) } ?
NSHipster
Swift Property Wrappers
Swift property wrappers go a long way to making SwiftUI possible, but they may play an even more important role in shaping the future of the language as a whole.
2 Replies
Valorie
Valorie15mo ago
That must be a typo in the article. $name produces an error and you’re right that _name works and makes sense.
kamilczajka
kamilczajkaOP15mo ago
Thanks @Valorie
Want results from more Discord servers?
Add your server