In Swift, we must specify the type for Character variables(because there’s no Character literals):
let eol: Character = “\n”
it's not a big problem, but, Is it the unique case that can’t use type infer?
In Swift, we must specify the type for Character variables(because there’s no Character literals):
let eol: Character = “\n”
it's not a big problem, but, Is it the unique case that can’t use type infer?
Well, what if you wanted to create a String with one character? There’s no
way to differentiate.
On Jul 7, 2016, at 02:35, 王 黎明 via swift-users [swift-users@swift.org](mailto:swift-users@swift.org) wrote: >
In Swift, we must specify the type for Character variables(because there’s
no Character literals):
let eol: Character = “\n”
it's not a big problem, but, Is it the unique case that can’t use type
infer?
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-Saagar Jha
Of course not. Struct Set is commonly known.
let foo:Set = [1,2,3,4]
let bar:Set<Int> = [1,2,3,4]
If you do not specify them as Sets, foo and bar will be Arrays.
Zhaoxin
On Thu, Jul 7, 2016 at 5:35 PM, 王 黎明 <swift-users@swift.org> wrote:
In Swift, we must specify the type for Character variables(because there’s
no Character literals):let eol: Character = “\n”
it's not a big problem, but, Is it the unique case that can’t use type
infer?_______________________________________________
swift-users mailing list
swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users
Well, what if you wanted to create a String with one character? There’s no way to differentiate.
That hardly seems like the justification. In that case, you'd specify the type:
let s: String = '\n'
On Jul 10, 2016, at 11:44 , Saagar Jha via swift-users <swift-users@swift.org> wrote:
> On Jul 7, 2016, at 02:35, 王 黎明 via swift-users [swift-users@swift.org](mailto:swift-users@swift.org) wrote: >
In Swift, we must specify the type for Character variables(because there’s no Character literals):
let eol: Character = “\n”
it's not a big problem, but, Is it the unique case that can’t use type infer?
swift-users mailing list swift-users@swift.org https://lists.swift.org/mailman/listinfo/swift-users
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-Saagar Jha
_______________________________________________
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Rick Mann
rmann@latencyzero.com
Yes, but then a similar complaint could be made that I wanted a String this time. The only real way to tell the difference is to use something like single quotes.
On Jul 10, 2016, at 14:03, Rick Mann via swift-users <swift-users@swift.org> wrote:
On Jul 10, 2016, at 11:44 , Saagar Jha via swift-users <swift-users@swift.org> wrote:
Well, what if you wanted to create a String with one character? There’s no way to differentiate.
That hardly seems like the justification. In that case, you'd specify the type:
let s: String = '\n'
On Jul 7, 2016, at 02:35, 王 黎明 via swift-users [swift-users@swift.org](mailto:swift-users@swift.org) wrote: >
In Swift, we must specify the type for Character variables(because there’s no Character literals):
let eol: Character = “\n”
it's not a big problem, but, Is it the unique case that can’t use type infer?
swift-users mailing list swift-users@swift.org https://lists.swift.org/mailman/listinfo/swift-users
--
-Saagar Jha
_______________________________________________
swift-users mailing list
swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users--
Rick Mann
rmann@latencyzero.com_______________________________________________
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swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users
In Ruby they have special literals, like so:
So this would be an array of chars
%c{a b c d e f g}
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On 11 July 2016 at 05:56, Saagar Jha via swift-users <swift-users@swift.org> wrote:
Yes, but then a similar complaint could be made that I *wanted* a String
this time. The only real way to tell the difference is to use something
like single quotes.On Jul 10, 2016, at 14:03, Rick Mann via swift-users < > swift-users@swift.org> wrote:
On Jul 10, 2016, at 11:44 , Saagar Jha via swift-users < > swift-users@swift.org> wrote:
Well, what if you wanted to create a String with one character? There’s no
way to differentiate.That hardly seems like the justification. In that case, you'd specify the
type:let s: String = '\n'
On Jul 7, 2016, at 02:35, 王 黎明 via swift-users [swift-users@swift.org]( > mailto:swift-users@swift.org <swift-users@swift.org>) wrote: >
In Swift, we must specify the type for Character variables(because there’s
no Character literals):let eol: Character = “\n”
it's not a big problem, but, Is it the unique case that can’t use type
infer?swift-users mailing list swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users--
-Saagar Jha
_______________________________________________
swift-users mailing list
swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users--
Rick Mann
rmann@latencyzero.com_______________________________________________
swift-users mailing list
swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users_______________________________________________
swift-users mailing list
swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users