blocks|key|127718|text|因为字符有一个有限的，有限的(不是太宽的)范围，你可以只测试哪些字符是给定字符集的成员(蛮力)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|127719|//+this+doesn't+seem+to+be+available
#define+UNICHAR_MAX+(1ull+<<+(CHAR_BIT+*+sizeof(unichar)))

NSData+*data+=+[[NSCharacterSet+uppercaseLetterCharacterSet]+bitmapRepresentation];
uint8_t+*ptr+=+[data+bytes];
NSMutableArray+*allCharsInSet+=+[NSMutableArray+array];
//+following+from+Apple's+sample+code
for+(unichar+i+=+0;+i+<+UNICHAR_MAX;+i%2B%2B)+{
++++if+(ptr[i+>>+3]+&+(1u+<<+(i+&+7)))+{
++++++++[allCharsInSet+addObject:[NSString+stringWithCharacters:&i+length:1]];
++++}
}|code-block|syntax|javascript|127720|备注:由于unichar的大小和bitmapRepresentation中附加段的结构，此解决方案仅适用于字符<=+0xFFFF，不适用于更高的平面。|127721|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Since characters have a limited, finite (and not too wide) range, you can just test which characters are members of a given character set (brute force):

<pre><code>// this doesn't seem to be available
#define UNICHAR_MAX (1ull &lt;&lt; (CHAR_BIT * sizeof(unichar)))

NSData *data = [[NSCharacterSet uppercaseLetterCharacterSet] bitmapRepresentation];
uint8_t *ptr = [data bytes];
NSMutableArray *allCharsInSet = [NSMutableArray array];
// following from Apple's sample code
for (unichar i = 0; i &lt; UNICHAR_MAX; i++) {
 if (ptr[i &gt;&gt; 3] &amp; (1u &lt;&lt; (i &amp; 7))) {
 [allCharsInSet addObject:[NSString stringWithCharacters:&amp;i length:1]];
 }
}
</code></pre>

Remark: Due to the size of a unichar and the structure of the additional segments in bitmapRepresentation, this solution only works for characters &lt;= 0xFFFF and is not suitable for higher planes.

blocks|key|345433|text|我创建了Martin+R算法的Swift+(v2.1)版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|345434|let+charset+=+NSCharacterSet.URLPathAllowedCharacterSet();

for+var+plane+:+UInt8+in+0...16+{
++++if+charset.hasMemberInPlane(+plane+)+{
++++++++var+c+:+UTF32Char;

++++++++for+var+c+:+UInt32+=+UInt32(+plane+)+<<+16;+c+<+(UInt32(plane)%2B1)+<<+16;+c%2B%2B+{
++++++++++++if+charset.longCharacterIsMember(c)+{
++++++++++++++++var+c1+=+c.littleEndian+//+To+make+it+byte-order+safe
++++++++++++++++let+s+=+NSString(bytes:+&c1,+length:+4,+encoding:+NSUTF32LittleEndianStringEncoding);
++++++++++++++++NSLog("Char:+\(s)");
++++++++++++}
++++++++}
++++}
}|code-block|syntax|javascript|345435|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I created a Swift (v2.1) version of Martin R's algorithm:

<pre><code>let charset = NSCharacterSet.URLPathAllowedCharacterSet();

for var plane : UInt8 in 0...16 {
 if charset.hasMemberInPlane( plane ) {
 var c : UTF32Char;

 for var c : UInt32 = UInt32( plane ) &lt;&lt; 16; c &lt; (UInt32(plane)+1) &lt;&lt; 16; c++ {
 if charset.longCharacterIsMember(c) {
 var c1 = c.littleEndian // To make it byte-order safe
 let s = NSString(bytes: &amp;c1, length: 4, encoding: NSUTF32LittleEndianStringEncoding);
 NSLog("Char: \(s)");
 }
 }
 }
}
</code></pre>

blocks|key|127809|text|这是使用更多的swift+for+swift来完成的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|127810|let+characters+=+NSCharacterSet.uppercaseLetterCharacterSet()
var+array++++++=+[String]()

for+plane:+UInt8+in+0...16+where+characters.hasMemberInPlane(plane)+{

++for+character:+UTF32Char+in+UInt32(plane)+<<+16..<(UInt32(plane)+%2B+1)+<<+16+where+characters.longCharacterIsMember(character)+{

++++var+endian+=+character.littleEndian
++++let+string+=+NSString(bytes:+&endian,+length:+4,+encoding:+NSUTF32LittleEndianStringEncoding)+as!+String

++++array.append(string)

++}

}

print(array)|code-block|syntax|javascript|127811|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is done using a little more of swift for swift.

<pre><code>let characters = NSCharacterSet.uppercaseLetterCharacterSet()
var array = [String]()

for plane: UInt8 in 0...16 where characters.hasMemberInPlane(plane) {

 for character: UTF32Char in UInt32(plane) &lt;&lt; 16..&lt;(UInt32(plane) + 1) &lt;&lt; 16 where characters.longCharacterIsMember(character) {

 var endian = character.littleEndian
 let string = NSString(bytes: &amp;endian, length: 4, encoding: NSUTF32LittleEndianStringEncoding) as! String

 array.append(string)

 }

}

print(array)
</code></pre>

blocks|key|345543|text|对于拉丁字母表中的A-Z+(没有希腊语，或者变音符号，或者其他不是这个人所要求的东西)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|345544|for+plane:+UInt8+in+0...16+where+characters.hasMemberInPlane(plane)+{
++++i+=+0
++++for+character:+UTF32Char+in+UInt32(plane)+<<+16...(UInt32(plane)+%2B+1)+<<+16+where+characters.longCharacterIsMember(character)+{
++++++++var+endian+=+character.littleEndian
++++++++let+string+=+NSString(bytes:+&endian,+length:+4,+encoding:+NSUTF32LittleEndianStringEncoding)+as!+String
++++++++array.append(string)
++++++++if(array.count+==+26)+{
++++++++++++break
++++++++}
++++}
++++if(array.count+==+26)+{
++++++++break
++++}
}|code-block|syntax|javascript|345545|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

For just A-Z of the Latin alphabet (nothing with Greek, or diacritical marks, or other things that were not what the guy asked for):

<pre><code>for plane: UInt8 in 0...16 where characters.hasMemberInPlane(plane) {
 i = 0
 for character: UTF32Char in UInt32(plane) &lt;&lt; 16...(UInt32(plane) + 1) &lt;&lt; 16 where characters.longCharacterIsMember(character) {
 var endian = character.littleEndian
 let string = NSString(bytes: &amp;endian, length: 4, encoding: NSUTF32LittleEndianStringEncoding) as! String
 array.append(string)
 if(array.count == 26) {
 break
 }
 }
 if(array.count == 26) {
 break
 }
}
</code></pre>

blocks|key|345651|text|我发现Martin+R的解决方案对我来说太慢了，所以我用另一种方法解决了这个问题，使用了CharacterSet的bitmapRepresentation属性。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|345652|根据我的基准测试，这要快得多：|345653|var+ranges+=+[CountableClosedRange<UInt32>]()
let+bitmap:+Data+=+characterSet.bitmapRepresentation
var+first:+UInt32?,+last:+UInt32?
var+plane+=+0,+nextPlane+=+8192
for+(j,+byte)+in+bitmap.enumerated()+where+byte+!=+0+{
++++if+j+==+nextPlane+{
++++++++plane+%2B=+1
++++++++nextPlane+%2B=+8193
++++++++continue
++++}
++++for+i+in+0+..<+8+where+byte+&+1+<<+i+!=+0+{
++++++++let+codePoint+=+UInt32(j+-+plane)+*+8+%2B+UInt32(i)
++++++++if+let+_last+=+last,+codePoint+==+_last+%2B+1+{
++++++++++++last+=+codePoint
++++++++}+else+{
++++++++++++if+let+first+=+first,+let+last+=+last+{
++++++++++++++++ranges.append(first+...+last)
++++++++++++}
++++++++++++first+=+codePoint
++++++++++++last+=+codePoint
++++++++}
++++}
}
if+let+first+=+first,+let+last+=+last+{
++++ranges.append(first+...+last)
}
return+ranges|code-block|syntax|javascript|345654|这个解决方案返回一个codePoint范围数组，但是您可以很容易地调整它以返回单个字符或字符串等。|345655|entityMap^0|18|C|1L|K|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

I found Martin R's solution to be too slow for my purposes, so I solved it another way using <code>CharacterSet</code>'s <code>bitmapRepresentation</code> property.

This is significantly faster according to my benchmarks:

<pre><code>var ranges = [CountableClosedRange&lt;UInt32&gt;]()
let bitmap: Data = characterSet.bitmapRepresentation
var first: UInt32?, last: UInt32?
var plane = 0, nextPlane = 8192
for (j, byte) in bitmap.enumerated() where byte != 0 {
 if j == nextPlane {
 plane += 1
 nextPlane += 8193
 continue
 }
 for i in 0 ..&lt; 8 where byte &amp; 1 &lt;&lt; i != 0 {
 let codePoint = UInt32(j - plane) * 8 + UInt32(i)
 if let _last = last, codePoint == _last + 1 {
 last = codePoint
 } else {
 if let first = first, let last = last {
 ranges.append(first ... last)
 }
 first = codePoint
 last = codePoint
 }
 }
}
if let first = first, let last = last {
 ranges.append(first ... last)
}
return ranges
</code></pre>

This solution returns an array of codePoint ranges, but you can easily adapt it to return individual characters or strings, etc.

blocks|key|345692|text|受Satachito+answer的启发，这里有一种使用bitmapRepresentation从CharacterSet生成数组的高效方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|345693|extension+CharacterSet+{
++++func+characters()+->+[Character]+{
++++++++//+A+Unicode+scalar+is+any+Unicode+code+point+in+the+range+U%2B0000+to+U%2BD7FF+inclusive+or+U%2BE000+to+U%2B10FFFF+inclusive.
++++++++return+codePoints().compactMap+{+UnicodeScalar($0)+}.map+{+Character($0)+}
++++}

++++func+codePoints()+->+[Int]+{
++++++++var+result:+[Int]+=+[]
++++++++var+plane+=+0
++++++++//+following+documentation+at+https://developer.apple.com/documentation/foundation/nscharacterset/1417719-bitmaprepresentation
++++++++for+(i,+w)+in+bitmapRepresentation.enumerated()+{
++++++++++++let+k+=+i+%25+0x2001
++++++++++++if+k+==+0x2000+{
++++++++++++++++//+plane+index+byte
++++++++++++++++plane+=+Int(w)+<<+13
++++++++++++++++continue
++++++++++++}
++++++++++++let+base+=+(plane+%2B+k)+<<+3
++++++++++++for+j+in+0+..<+8+where+w+&+1+<<+j+!=+0+{
++++++++++++++++result.append(base+%2B+j)
++++++++++++}
++++++++}
++++++++return+result
++++}
}|code-block|syntax|javascript|345694|uppercaseLetters示例|345695|let+charset+=+CharacterSet.uppercaseLetters
let+chars+=+charset.characters()
print(chars.count)+//+1733
print(chars)+//+["A",+"B",+"C",+...+"]|345696|不连续平面的示例|345697|let+charset+=+CharacterSet(charactersIn:+"")
let+codePoints+=+charset.codePoints()
print(codePoints)+//+[120488,+837521]|345698|性能|345699|非常好:这个使用bitmapRepresentation发布的解决方案似乎比Martin+R使用contains的解决方案或Oliver+Atkinson使用longCharacterIsMember的解决方案快3到10倍。|345700|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/51200984/1033581|1|https://developer.apple.com/documentation/foundation/characterset/1779849-bitmaprepresentation|2|https://developer.apple.com/documentation/foundation/characterset/2908835-contains|3|https://developer.apple.com/documentation/foundation/nscharacterset/1409757-longcharacterismember^0|S|K|1|G|0|0|0|0|0|0|0|0|8|K|1C|8|27|L|8|K|1|1C|8|2|27|L|3|0^^$0|@$1|2|3|4|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|F|3|G|5|H|7|1G|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1H|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1I|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1J|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1K|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1L|8|@]|D|@]|E|$]]|$1|U|3|V|5|6|7|1M|8|@$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]|$9|1R|A|1S|B|C]]|D|@$9|1T|A|1U|1|1V]|$9|1W|A|1X|1|1Y]|$9|1Z|A|20|1|21]]|E|$]]|$1|W|3|-4|5|6|7|22|8|@]|D|@]|E|$]]]|X|$Y|$5|Z|10|11|E|$12|13]]|14|$5|Z|10|11|E|$12|15]]|16|$5|Z|10|11|E|$12|17]]|18|$5|Z|10|11|E|$12|19]]]]

Inspired by <a href="https://stackoverflow.com/a/51200984/1033581">Satachito answer</a>, here is a performant way to make an Array from CharacterSet using <code>bitmapRepresentation</code>:

<pre><code>extension CharacterSet {
 func characters() -&gt; [Character] {
 // A Unicode scalar is any Unicode code point in the range U+0000 to U+D7FF inclusive or U+E000 to U+10FFFF inclusive.
 return codePoints().compactMap { UnicodeScalar($0) }.map { Character($0) }
 }

 func codePoints() -&gt; [Int] {
 var result: [Int] = []
 var plane = 0
 // following documentation at https://developer.apple.com/documentation/foundation/nscharacterset/1417719-bitmaprepresentation
 for (i, w) in bitmapRepresentation.enumerated() {
 let k = i % 0x2001
 if k == 0x2000 {
 // plane index byte
 plane = Int(w) &lt;&lt; 13
 continue
 }
 let base = (plane + k) &lt;&lt; 3
 for j in 0 ..&lt; 8 where w &amp; 1 &lt;&lt; j != 0 {
 result.append(base + j)
 }
 }
 return result
 }
}
</code></pre>

<h3>Example for uppercaseLetters</h3>

<pre><code>let charset = CharacterSet.uppercaseLetters
let chars = charset.characters()
print(chars.count) // 1733
print(chars) // ["A", "B", "C", ... "]
</code></pre>

<h3>Example for discontinuous planes</h3>

<pre><code>let charset = CharacterSet(charactersIn: "")
let codePoints = charset.codePoints()
print(codePoints) // [120488, 837521]
</code></pre>

<h3>Performances</h3>

Very good: this solution built in release with <a href="https://developer.apple.com/documentation/foundation/characterset/1779849-bitmaprepresentation" rel="noreferrer"><code>bitmapRepresentation</code></a> seems 3 to 10 times faster than Martin R's solution with <a href="https://developer.apple.com/documentation/foundation/characterset/2908835-contains" rel="noreferrer"><code>contains</code></a> or Oliver Atkinson's solution with <a href="https://developer.apple.com/documentation/foundation/nscharacterset/1409757-longcharacterismember" rel="noreferrer"><code>longCharacterIsMember</code></a>.

blocks|key|128051|text|当然，您可以使用CharacterSet创建字符集和字母表，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|128052|var+smallEmojiCharacterSet+=+CharacterSet(charactersIn:++Unicode.Scalar("")...Unicode.Scalar(""))|code-block|syntax|javascript|128053|问题是CharacterSet不是Set+(尽管它符合SetAlgebra)，而是一个unicode字符集。这导致了获取其所有字符的序列的问题，以便将其转换为Array、Set或String。我have+found+a+solution，但有一个更好的存在。实际上，你想要的是从一个字符到另一个字符，有一个范围"a"..."z“。这在标量级别上并不难做到。在Character级别，有更多的注意事项需要考虑。|128054|extension+Unicode.Scalar:+Strideable+{
++++public+typealias+Stride+=+Int

++++public+func+distance(to+other:+Unicode.Scalar)+->+Int+{
++++++++return+Int(other.value)+-+Int(self.value)
++++}

++++public+func+advanced(by+n:+Int)+->+Unicode.Scalar+{
++++++++return+Unicode.Scalar(UInt32(Int(value)+%2B+n))!
++++}
}


let+alphabetScalarRange+=+(Unicode.Scalar("a")...Unicode.Scalar("z"))//+ClosedRange<Unicode.Scalar>

let+alphabetCharactersArr+=+Array(alphabetScalarRange.map(Character.init))+//+Array+of+Characters+from+range
let+alphabetStringsArr+=+Array(alphabetScalarRange.map(String.init))+//+Array+of+Strings+from+range
let+alphabetString+=+alphabetStringsArr.joined()+//+String+(collection+of+characters)+from+range
//+or+simply
let+uppercasedAlphabetString+=++(("A"+as+Unicode.Scalar)..."Z").reduce("")+{+(r,+us)+->+String+in
++++r+%2B+String(us)
}|128055|如果你认为做一个扩展是一件过分的事情|128056|let+alphabetScalarValueRange+=+(Unicode.Scalar("a").value...Unicode.Scalar("z").value)
let+alphabetStringsArr2+=+Array(alphabetScalarValueRange.compactMap{+Unicode.Scalar($0)?.escaped(asASCII:+false)+})
let+alphabetString2+=+alphabetScalarValueRange.compactMap({+Unicode.Scalar($0)?.escaped(asASCII:+false)+}).joined(separator:+",+")|128057|但是要小心:字符可以由几个标量组成。|128058|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/55473051/converting-a-characterset-to-a-string-or-to-an-array-or-set-of-characters^0|8|C|0|0|3|C|H|3|R|A|27|5|2D|3|2H|6|4Z|9|2P|L|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]]|D|@$9|1L|A|1M|1|1N]]|E|$]]|$1|M|3|N|5|H|7|1O|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1P|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1Q|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1R|8|@]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1S|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]]]

You can of course create sets of characters and alphabets using <code>CharacterSet</code> like this:

<pre><code>var smallEmojiCharacterSet = CharacterSet(charactersIn: Unicode.Scalar("")...Unicode.Scalar(""))
</code></pre>

The problem is that <code>CharacterSet</code> is NOT a <code>Set</code> (though it conforms to <code>SetAlgebra</code>), it is rather a unicode character set . This causes the problem of getting a sequence of all it's characters, to convert it to <code>Array</code>, <code>Set</code> or a <code>String</code>. I <a href="https://stackoverflow.com/questions/55473051/converting-a-characterset-to-a-string-or-to-an-array-or-set-of-characters">have found a solution</a>, but a better one exists. 
Actually, what you want is to stride from character to character, to have a range "a"..."z".
It is not hard to do at the scalar level. At <code>Character</code> level there are more caveats to consider.

<pre><code>extension Unicode.Scalar: Strideable {
 public typealias Stride = Int

 public func distance(to other: Unicode.Scalar) -&gt; Int {
 return Int(other.value) - Int(self.value)
 }

 public func advanced(by n: Int) -&gt; Unicode.Scalar {
 return Unicode.Scalar(UInt32(Int(value) + n))!
 }
}


let alphabetScalarRange = (Unicode.Scalar("a")...Unicode.Scalar("z"))// ClosedRange&lt;Unicode.Scalar&gt;

let alphabetCharactersArr = Array(alphabetScalarRange.map(Character.init)) // Array of Characters from range
let alphabetStringsArr = Array(alphabetScalarRange.map(String.init)) // Array of Strings from range
let alphabetString = alphabetStringsArr.joined() // String (collection of characters) from range
// or simply
let uppercasedAlphabetString = (("A" as Unicode.Scalar)..."Z").reduce("") { (r, us) -&gt; String in
 r + String(us)
}
</code></pre>

If you think making an extension is an overkill

<pre><code>let alphabetScalarValueRange = (Unicode.Scalar("a").value...Unicode.Scalar("z").value)
let alphabetStringsArr2 = Array(alphabetScalarValueRange.compactMap{ Unicode.Scalar($0)?.escaped(asASCII: false) })
let alphabetString2 = alphabetScalarValueRange.compactMap({ Unicode.Scalar($0)?.escaped(asASCII: false) }).joined(separator: ", ")
</code></pre>

But be careful: Characters can consist of several scalars.

Currently I am able to make array of Alphabets like below

<pre><code>[[NSArray alloc]initWithObjects:@"A",@"B",@"C",@"D",@"E",@"F",@"G",@"H",@"I",@"J",@"K",@"L",@"M",@"N",@"O",@"P",@"Q",@"R",@"S",@"T",@"U",@"V",@"W",@"X",@"Y",@"Z",nil];
</code></pre>

Knowing that is available over

<pre><code>[NSCharacterSet uppercaseLetterCharacterSet]
</code></pre>

How to make an array out of it?

NSArray from NSCharacterSet

Swift

目前我可以像下面这样做字母表的数组[[NSArray alloc]initWithObjects:@"A",@"B",@"C",@"D",@"E",@"F",@"G",@"H",@"I",@"J",@"K",@"L",@"M",@"N",@"O",@"P",@"Q",@"R",@"S",@"T",@"U",@"V",@...

问来自NSCharacterSet的NSArray
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