A common issue many of the developers might face while building an app with localization (especially when it involves Arabic or an RTL-supported language), is that the search would not result as expected. An example of the issue is:
print(listOfName.contains(InputName))); //prints false
To overcome this issue, I tried comparing the encoded search strings (in both languages) and encoded result strings then only realized that somehow some special characters had been added. For my instance, the characters were RTL[226, 128, 143] and LTR[226, 128, 142]. Before I actually encoded the strings, both search and result were identical or equals to the same. After knowing the extra added characters, I did the following:
var InputNameEncode = InputName.encode
var rightToLeftMark = utf8.decode([226, 128, 143]);
var leftToRightMark = utf8.decode([226, 128, 142]);
InputNameEncode = InputNameEncode.replaceAll(rightToLeftMark, "");
InputNameEncode = InputNameEncode.replaceAll(leftToRightMark, "");
inputName = InputNameEncode.decode
print(listOfName.contains(InputName))); //prints true
As mentioned earlier, in my case the extra characters were RTL and LTR. In your case, you may find something entirely different.
You can know what each encoded set of characters represents on this page: https://www.utf8-chartable.de/unicode-utf8-table.pl?start=8192&number=128&utf8=dec
Flutter Version: 1.22.6
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So, for example the character 김 is made up of ㄱ, ㅣ and ㅁ. I need to split the Korean word into it's components to get the resulting 3 characters.
I tried by doing the following but it doesn't seem to output it correctly:
let str = "김"
let utf8 = str.utf8
let first:UInt8 = utf8.first!
let char = Character(UnicodeScalar(first))
The problem is, that that code returns ê, when it should be returning ㄱ.
You need to use the decomposedStringWithCompatibilityMapping string to get the unicode scalar values and then use those scalar values to get the characters. Something below,
let string = "김"
for scalar in string.decomposedStringWithCompatibilityMapping.unicodeScalars {
print("\(scalar) ")
}
Output:
ᄀ
ᅵ
ᆷ
You can create list of character strings as,
let chars = string.decomposedStringWithCompatibilityMapping.unicodeScalars.map { String($0) }
print(chars)
// ["ᄀ", "ᅵ", "ᆷ"]
Korean related info in Apple docs
Extended grapheme clusters are a flexible way to represent many
complex script characters as a single Character value. For example,
Hangul syllables from the Korean alphabet can be represented as either
a precomposed or decomposed sequence. Both of these representations
qualify as a single Character value in Swift:
let precomposed: Character = "\u{D55C}" // 한
let decomposed: Character = "\u{1112}\u{1161}\u{11AB}" // ᄒ, ᅡ, ᆫ
// precomposed is 한, decomposed is 한
I am dealing with strings containing \r\n with Swift 4.2. I ran into kind of strange behavior of Swift index, it appears \r\n will be treated as one character instead of two by Swift indexing methods. I wrote a piece of code to present this behavior:
var text = "ABC\r\n\r\nDEF"
func printChar(_ lower: Int, _ upper: Int) {
let start = text.index(text.startIndex, offsetBy: lower)
let end = text.index(text.startIndex, offsetBy: upper)
print("\"" + text[start..<end] + "\"")
}
printChar(0, 1) // "A"
printChar(1, 2) // "B"
printChar(2, 3) // "C"
printChar(3, 4) // new line
printChar(4, 5) // new line (okay, what's going on here?)
printChar(5, 6) // "D"
printChar(6, 7) // "E"
printChar(7, 8) // "F"
The print result will be
"A"
"B"
"C"
"
"
"
"
"D"
"E"
"F"
Any idea why it's like this?
TLDR: \r\n is a grapheme cluster and is treated as a single Character in Swift because Unicode.
Swift treats \r\n as one Character.
Objective-C NSString treats it as two characters (in terms of the result from length).
On the swift-users forum someone wrote:
– "\r\n" is a single Character. Is this the correct behaviour?
– Yes, a Character corresponds to a Unicode grapheme cluster, and "\r\n" is considered a single grapheme cluster.
And the subsequent response posted a link to Unicode documentation, check out this table which officially states CRLF is a grapheme cluster.
Take a look at the Apple documentation on Characters and Grapheme Clusters.
It's common to think of a string as a sequence of characters, but when working with NSString objects, or with Unicode strings in general, in most cases it is better to deal with substrings rather than with individual characters. The reason for this is that what the user perceives as a character in text may in many cases be represented by multiple characters in the string.
The Swift documentation on Strings and Characters is also worth reading.
This overview from objc.io is interesting as well.
NSString represents UTF-16-encoded text. Length, indices, and ranges are all based on UTF-16 code units.
Another example of this is an emoji like 👍🏻. This single character is actually %uD83D%uDC4D%uD83C%uDFFB, four different unicode scalars. But if you called count on a string with just that emoji you'd (correctly) get 1.
If you wanted to see the scalars you could iterate them as follows:
for scalar in text.unicodeScalars {
print("\(scalar.value) ", terminator: "")
}
Which for "\r\n" would give you 13 10
In the Swift documentation you'll find why NSString is different:
The count of the characters returned by the count property isn’t always the same as the length property of an NSString that contains the same characters. The length of an NSString is based on the number of 16-bit code units within the string’s UTF-16 representation and not the number of Unicode extended grapheme clusters within the string.
Thus this isn't really "strange" behaviour of Swift string indexing, but rather a result of how Unicode treats these characters and how String in Swift is designed. Swift string indexing goes by Character and \r\n is a single Character.
I have the code snippet below (as suggested in this previous Stack Overflow answer ... Deleting all special characters from a string in progress 4GL) which is attempting to remove all extended characters from a string so that I may transmit it to a customer's system which will not accept any extended characters.
do v-int = 128 to 255:
assign v-string = replace(v-string,chr(v-int),"").
end.
It is working perfectly with one exception (which makes me fear there may be others I have not caught). When it gets to 255, it will replace all 'y's in the string.
If I do the following ...
display chr(255) = chr(121). /* 121 is asc code of y */
I get true as the result.
And therefore, if I do the following ...
display replace("This is really strange",chr(255),"").
I get the following result:
This is reall strange
I have verified that 'y' is the only character affected by running the following:
def var v-string as char init "abcdefghijklmnopqrstuvwxyz".
def var v-int as int.
do v-int = 128 to 255:
assign v-string = replace(v-string,chr(v-int),"").
end.
display v-string.
Which results in the following:
abcdefghijklmnopqrstuvwxz
I know I can fix this by removing 255 from the range but I would like to understand why this is happening.
Is this a character collation set issue or am I missing something simpler?
Thanks for any help!
This is a bug. Here's a Progress Knowledge Base article about it:
http://knowledgebase.progress.com/articles/Article/000046181
The workaround is to specify the codepage in the CHR() statement, like this:
CHR(255, "UTF-8", "1252")
Here it is in your example:
def var v-string as char init "abcdefghijklmnopqrstuvwxyz". def var v-int as int.
do v-int = 128 to 255:
assign v-string = replace(v-string, chr(v-int, "UTF-8", "1252"), "").
end.
display v-string.
You should now see the 'y' in the output.
This seems to be a bug!
The REPLACE() function returns an unexpected result when replacing character CHR(255) (ÿ) in a String.
The REPLACE() function modifies the value of the target character, but additionally it changes any occurrence of characters 'Y' and 'y' present in the String.
This behavior seems to affect only the character ÿ. Other characters are correctly changed by REPLACE().
Using default codepage ISO-8859-1
Link to knowledgebase
I figured out, that "ß" is converted to "SS" when using uppercased(). But I want compare if two strings are equal without being case sensitive.
So when comparing "gruß" with "GRUß" it should be the same as well as when comparing "gru" with "Gru".
There is no uppercase "ß" in german language! Because I don't know which other characters aren't available in the corresponding languages, I could not filter all the caracters, which have no 1:1 uppercased opponent.
What can I do?
Use caseInsensitiveCompare() instead of converting the strings
to upper or lowercase:
let s1 = "gruß"
let s2 = "GRUß"
let eq = s1.caseInsensitiveCompare(s2) == .orderedSame
print(eq) // true
This compares the strings in a case-insensitive way according to
the Unicode standard.
There is also localizedCaseInsensitiveCompare() which does
a comparison according to the current locale, and
s1.compare(s2, options: .caseInsensitive, locale: ...)
for a case-insensitive comparison according to an arbitrary given
locale.
Well, "GRUß" is non-sensical in the first place for the reasons you state. You can't throw arbitrary data at a computer and expect it to process it in a sane way :-) If you have to deal with invalid input, you should probably have a preprocessing phase which cleans up crap you know about.
Having said that, this works (Swift 3.0):
let grusz = "GRUß"
let gruss = "GRUSS"
if grusz.compare(gruss, options: .caseInsensitive) == .orderedSame {
print("MATCHES")
}
Swift seems to be trying to deprecate the notion of a string being composed of an array of atomic characters, which makes sense for many uses, but there's an awful lot of programming that involves picking through datastructures that are ASCII for all practical purposes: particularly with file I/O. The absence of a built in language feature to specify a character literal seems like a gaping hole, i.e. there is no analog of the C/Java/etc-esque:
String foo="a"
char bar='a'
This is rather inconvenient, because even if you convert your strings into arrays of characters, you can't do things like:
let ch:unichar = arrayOfCharacters[n]
if ch >= 'a' && ch <= 'z' {...whatever...}
One rather hacky workaround is to do something like this:
let LOWCASE_A = ("a" as NSString).characterAtIndex(0)
let LOWCASE_Z = ("z" as NSString).characterAtIndex(0)
if ch >= LOWCASE_A && ch <= LOWCASE_Z {...whatever...}
This works, but obviously it's pretty ugly. Does anyone have a better way?
Characters can be created from Strings as long as those Strings are only made up of a single character. And, since Character implements ExtendedGraphemeClusterLiteralConvertible, Swift will do this for you automatically on assignment. So, to create a Character in Swift, you can simply do something like:
let ch: Character = "a"
Then, you can use the contains method of an IntervalType (generated with the Range operators) to check if a character is within the range you're looking for:
if ("a"..."z").contains(ch) {
/* ... whatever ... */
}
Example:
let ch: Character = "m"
if ("a"..."z").contains(ch) {
println("yep")
} else {
println("nope")
}
Outputs:
yep
Update: As #MartinR pointed out, the ordering of Swift characters is based on Unicode Normalization Form D which is not in the same order as ASCII character codes. In your specific case, there are more characters between a and z than in straight ASCII (ä for example). See #MartinR's answer here for more info.
If you need to check if a character is in between two ASCII character codes, then you may need to do something like your original workaround. However, you'll also have to convert ch to an unichar and not a Character for it to work (see this question for more info on Character vs unichar):
let a_code = ("a" as NSString).characterAtIndex(0)
let z_code = ("z" as NSString).characterAtIndex(0)
let ch_code = (String(ch) as NSString).characterAtIndex(0)
if (a_code...z_code).contains(ch_code) {
println("yep")
} else {
println("nope")
}
Or, the even more verbose way without using NSString:
let startCharScalars = "a".unicodeScalars
let startCode = startCharScalars[startCharScalars.startIndex]
let endCharScalars = "z".unicodeScalars
let endCode = endCharScalars[endCharScalars.startIndex]
let chScalars = String(ch).unicodeScalars
let chCode = chScalars[chScalars.startIndex]
if (startCode...endCode).contains(chCode) {
println("yep")
} else {
println("nope")
}
Note: Both of those examples only work if the character only contains a single code point, but, as long as we're limited to ASCII, that shouldn't be a problem.
If you need C-style ASCII literals, you can just do this:
let chr = UInt8(ascii:"A") // == UInt8( 0x41 )
Or if you need 32-bit Unicode literals you can do this:
let unichr1 = UnicodeScalar("A").value // == UInt32( 0x41 )
let unichr2 = UnicodeScalar("é").value // == UInt32( 0xe9 )
let unichr3 = UnicodeScalar("😀").value // == UInt32( 0x1f600 )
Or 16-bit:
let unichr1 = UInt16(UnicodeScalar("A").value) // == UInt16( 0x41 )
let unichr2 = UInt16(UnicodeScalar("é").value) // == UInt16( 0xe9 )
All of these initializers will be evaluated at compile time, so it really is using an immediate literal at the assembly instruction level.
The feature you want was proposed to be in Swift 5.1, but that proposal was rejected for a few reasons:
Ambiguity
The proposal as written, in the current Swift ecosystem, would have allowed for expressions like 'x' + 'y' == "xy", which was not intended (the proper syntax would be "x" + "y" == "xy").
Amalgamation
The proposal was two in one.
First, it proposed a way to introduce single-quote literals into the language.
Second, it proposed that these would be convertible to numerical types to deal with ASCII values and Unicode codepoints.
These are both good proposals, and it was recommended that this be split into two and re-proposed. Those follow-up proposals have not yet been formalized.
Disagreement
It never reached consensus whether the default type of 'x' would be a Character or a Unicode.Scalar. The proposal went with Character, citing the Principle of Least Surprise, despite this lack of consensus.
You can read the full rejection rationale here.
The syntax might/would look like this:
let myChar = 'f' // Type is Character, value is solely the unicode U+0066 LATIN SMALL LETTER F
let myInt8: Int8 = 'f' // Type is Int8, value is 102 (0x66)
let myUInt8Array: [UInt8] = [ 'a', 'b', '1', '2' ] // Type is [UInt8], value is [ 97, 98, 49, 50 ] ([ 0x61, 0x62, 0x31, 0x32 ])
switch someUInt8 {
case 'a' ... 'f': return "Lowercase hex letter"
case 'A' ... 'F': return "Uppercase hex letter"
case '0' ... '9': return "Hex digit"
default: return "Non-hex character"
}
It also looks like you can use the following syntax:
Character("a")
This will create a Character from the specified single character string.
I have only tested this in Swift 4 and Xcode 10.1
Why do I exhume 7 year old posts? Fun I guess? Seriously though, I think I can add to the discussion.
It is not a gaping hole, or rather, it is a deliberate gaping hole that explicitly discourages conflating a string of text with a sequence of ASCII bytes.
You absolutely can pick apart a String. A String implements BidirectionalCollection and has many ways to manipulate the atoms. See: https://developer.apple.com/documentation/swift/string.
But you have to get used to the more generalized notion of a String. It can be picked apart from the User perspective, which is a sequence of grapheme clusters, each (usually) which a visually separable appearance, or from the encoding perspective, which can be one of several (UTF32, UTF16, UTF8).
At the risk of overanalyzing the wording of your question:
A data structure is conceptual, and independent of encoding in storage
A data structure encoded as an ASCII string is just one kind of ASCII string
By design the encoding of ASCII values 0-127 will have an identical encoding in UTF-8, so loading that stream with a UTF8 API is fine
A data structure encoded as a string where fields of the structure have UTF-8 Unicode string values is not an ASCII string, but a UTF-8 string itself
A string is either ASCII-encoded or not; "for practical purposes" isn't a meaningful qualifier. A UTF-8 database field where 99.99% of the text falls in the ASCII range (where encodings will match), but occasionally doesn't, will present some nasty bug opportunities.
Instead of a terse and low-level equivalence of fixed-width integers and English-only text, Swift has a richer API that forces more explicit naming of the involved categories and entities. If you want to deal with ASCII, there's a name (method) for that, and if you want to deal with human sub-categories, there's a name for that, too, and they're totally independent of one another. There is a strong move away from ASCII and the English-centric string handling model of C. This is factual, not evangelizing, and it can present an irksome learning curve.
(This is aimed at new-comers, acknowledging the OP probably has years of experience with this now.)
For what you're trying to do there, consider:
let foo = "abcDeé#¶œŎO!##"
foo.forEach { c in
print((c.isASCII ? "\(c) is ascii with value \(c.asciiValue ?? 0); " : "\(c) is not ascii; ")
+ ((c.isLetter ? "\(c) is a letter" : "\(c) is not a letter")))
}
b is ascii with value 98; b is a letter
c is ascii with value 99; c is a letter
D is ascii with value 68; D is a letter
e is ascii with value 101; e is a letter
é is not ascii; é is a letter
# is ascii with value 64; # is not a letter
¶ is not ascii; ¶ is not a letter
œ is not ascii; œ is a letter
Ŏ is not ascii; Ŏ is a letter
O is ascii with value 79; O is a letter
! is ascii with value 33; ! is not a letter
# is ascii with value 64; # is not a letter
# is ascii with value 35; # is not a letter