I want to convert the index of a letter contained within a string to an integer value. Attempted to read the header files but I cannot find the type for Index
, although it appears to conform to protocol ForwardIndexType
with methods (e.g. distanceTo
).
我要将字符串中包含的字母的索引转换为整数值。试图读取头文件,但我找不到索引的类型,尽管它似乎与协议转发索引类型(例如distanceTo)一致。
var letters = "abcdefg"
let index = letters.characters.indexOf("c")!
// ERROR: Cannot invoke initializer for type 'Int' with an argument list of type '(String.CharacterView.Index)'
let intValue = Int(index) // I want the integer value of the index (e.g. 2)
Any help is appreciated.
任何帮助都是感激。
1 个解决方案
#1
40
You need to use the method distanceTo(index) in relation to the original string start index:
你需要使用方法distanceTo(index)相对于原始字符串start index:
let intValue = letters.startIndex.distanceTo(index)
You can also extend String with a method to return the first occurrence of a character in a string as follow:
您还可以使用方法扩展字符串,以返回字符串中第一个字符的出现,如下所示:
extension String {
func indexDistanceOfFirst(character character: Character) -> Int? {
guard let index = characters.indexOf(character) else { return nil }
return startIndex.distanceTo(index)
}
}
let letters = "abcdefg"
let char: Character = "c"
if let index = letters.indexDistanceOfFirst(character: char) {
print("character \(char) was found at position #\(index)") // "character c was found at position #2\n"
} else {
print("character \(char) was not found")
}
Xcode 8 • Swift 3
Xcode 8•Swift 3。
let letters = "abcdefg"
if let index = letters.characters.index(of: "c") {
let distance = letters.distance(from: letters.startIndex, to: index)
print("distance:", distance)
}
extension String {
func indexDistance(of character: Character) -> Int? {
guard let index = characters.index(of: character) else { return nil }
return distance(from: startIndex, to: index)
}
}
let letters = "abcdefg"
let char: Character = "c"
if let index = letters.indexDistance(of: char) {
print("character \(char) was found at position #\(index)") // "character c was found at position #2\n"
} else {
print("character \(char) was not found")
}
Xcode 9 • Swift 4
Xcode 9•Swift 4
let letters = "abcdefg"
if let index = letters.index(of: "c") {
let distance = letters.distance(from: letters.startIndex, to: index)
print("distance:", distance)
}
extension String {
func indexDistance(of character: Character) -> Int? {
guard let index = index(of: character) else { return nil }
return distance(from: startIndex, to: index)
}
}
Another possible approach in Swift 4 is to return the index encodedOffset
:
Swift 4的另一种可能方法是返回索引encodedOffset:
extension String {
func encodedOffset(of character: Character) -> Int? {
return index(of: character)?.encodedOffset
}
func encodedOffset(of string: String) -> Int? {
return range(of: string)?.lowerBound.encodedOffset
}
}
let letters = "abcdefg"
let char: Character = "c"
if let index = letters.encodedOffset(of: char) {
print("character \(char) was found at position #\(index)") // "character c was found at position #2\n"
} else {
print("character \(char) was not found")
}
#1
40
You need to use the method distanceTo(index) in relation to the original string start index:
你需要使用方法distanceTo(index)相对于原始字符串start index:
let intValue = letters.startIndex.distanceTo(index)
You can also extend String with a method to return the first occurrence of a character in a string as follow:
您还可以使用方法扩展字符串,以返回字符串中第一个字符的出现,如下所示:
extension String {
func indexDistanceOfFirst(character character: Character) -> Int? {
guard let index = characters.indexOf(character) else { return nil }
return startIndex.distanceTo(index)
}
}
let letters = "abcdefg"
let char: Character = "c"
if let index = letters.indexDistanceOfFirst(character: char) {
print("character \(char) was found at position #\(index)") // "character c was found at position #2\n"
} else {
print("character \(char) was not found")
}
Xcode 8 • Swift 3
Xcode 8•Swift 3。
let letters = "abcdefg"
if let index = letters.characters.index(of: "c") {
let distance = letters.distance(from: letters.startIndex, to: index)
print("distance:", distance)
}
extension String {
func indexDistance(of character: Character) -> Int? {
guard let index = characters.index(of: character) else { return nil }
return distance(from: startIndex, to: index)
}
}
let letters = "abcdefg"
let char: Character = "c"
if let index = letters.indexDistance(of: char) {
print("character \(char) was found at position #\(index)") // "character c was found at position #2\n"
} else {
print("character \(char) was not found")
}
Xcode 9 • Swift 4
Xcode 9•Swift 4
let letters = "abcdefg"
if let index = letters.index(of: "c") {
let distance = letters.distance(from: letters.startIndex, to: index)
print("distance:", distance)
}
extension String {
func indexDistance(of character: Character) -> Int? {
guard let index = index(of: character) else { return nil }
return distance(from: startIndex, to: index)
}
}
Another possible approach in Swift 4 is to return the index encodedOffset
:
Swift 4的另一种可能方法是返回索引encodedOffset:
extension String {
func encodedOffset(of character: Character) -> Int? {
return index(of: character)?.encodedOffset
}
func encodedOffset(of string: String) -> Int? {
return range(of: string)?.lowerBound.encodedOffset
}
}
let letters = "abcdefg"
let char: Character = "c"
if let index = letters.encodedOffset(of: char) {
print("character \(char) was found at position #\(index)") // "character c was found at position #2\n"
} else {
print("character \(char) was not found")
}