How do we find out the length (byte-wise) of a string [declared String type, not NSString] in Swift 2.2?
I know one way out is to use _bridgeToObejectiveC().length
Is there any other way out?
let str: String = "Hello, World"
print(str.characters.count) // 12
let str1: String = "Hello, World"
print(str1.endIndex) // 12
let str2 = "Hello, World"
NSString(string: str2).length //12
Refer String length in Swift 1.2 and Swift 2.0 and Get the length of a String
let str: String = "Hello, World"
print(str.characters.count) // 12
let byteLength = str.lengthOfBytesUsingEncoding(NSUTF8StringEncoding)
print("byte lenght: \(byteLength)")//12
let str2: String = "你好"
print(str2.characters.count) // 2
let byteLength2 = str.lengthOfBytesUsingEncoding(NSUTF8StringEncoding)
print("byte lenght: \(byteLength2)")//12
Try this code:
Swift
let test : String = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
let bytes : NSInteger = test.lengthOfBytesUsingEncoding(NSUTF8StringEncoding);
NSLog("%i bytes", bytes);
Objective C
refer this link:
What is the length in bytes of a NSString?
I think type casting is another easier way to use the methods, properties of NSString class.
print((str as NSString).length)
Related
Objective-C code:
NSString *str = #"hi";
NSString *strDigit = #"1934"; (or #"193" may be a 3 digit or 4 digit value)
[dayText appendFormat:#"%#%4s,str,[strDigit UTF8String]];
The Objective-C code handles the output string with current alignment when it appears with 3 or 4 digits as output. It is correctly aligning to left and doesn't matter how much digits it is. Any one know how to handle this in Swift?
In Swift I tried with below code and the string is not adjusting the alignment according to the number of digits.
textForTrip += "\(str) \(String(format:"%4s", (strDigit.utf8))"
The %s format expects a pointer to a (NULL-terminated) C string
as argument, that can be obtained with the withCString method.
This would produce the same output as your Objective-C code:
let str = "Hi"
let strDigit = "193"
let text = strDigit.withCString {
String(format: "%#%4s", str, $0)
}
print(text)
It becomes easier if you store the number as integer instead of a
string:
let str = "Hi"
let number = 934
let text = String(format: "%#%4d", str, number)
print(text)
Try this below approach, that might help you
let strDigit = "\("1934".utf8)" //(or #"193" may be a 3 digit or 4 digit value)
var dayText = "Hello, good morning."
dayText += "\(strDigit.prefix(3))"
I have a problem I couldn't find a solution to.
I have a string variable holding the unicode "1f44d" and I want to convert it to a unicode character 👍.
Usually one would do something like this:
println("\u{1f44d}") // 👍
Here is what I mean:
let charAsString = "1f44d" // code in variable
println("\u{\(charAsString)}") // not working
I have tried several other ways but somehow the workings behind this magic stay hidden for me.
One should imagine the value of charAsString coming from an API call or from another object.
One possible solution (explanations "inline"):
let charAsString = "1f44d"
// Convert hex string to numeric value first:
var charCode : UInt32 = 0
let scanner = NSScanner(string: charAsString)
if scanner.scanHexInt(&charCode) {
// Create string from Unicode code point:
let str = String(UnicodeScalar(charCode))
println(str) // 👍
} else {
println("invalid input")
}
Slightly simpler with Swift 2:
let charAsString = "1f44d"
// Convert hex string to numeric value first:
if let charCode = UInt32(charAsString, radix: 16) {
// Create string from Unicode code point:
let str = String(UnicodeScalar(charCode))
print(str) // 👍
} else {
print("invalid input")
}
Note also that not all code points are valid Unicode scalars,
compare Validate Unicode code point in Swift.
Update for Swift 3:
public init?(_ v: UInt32)
is now a failable initializer of UnicodeScalar and checks if the
given numeric input is a valid Unicode scalar value:
let charAsString = "1f44d"
// Convert hex string to numeric value first:
if let charCode = UInt32(charAsString, radix: 16),
let unicode = UnicodeScalar(charCode) {
// Create string from Unicode code point:
let str = String(unicode)
print(str) // 👍
} else {
print("invalid input")
}
This can be done in two steps:
convert charAsString to Int code
convert code to unicode character
Second step can be done e.g. like this
var code = 0x1f44d
var scalar = UnicodeScalar(code)
var string = "\(scalar)"
As for first the step, see here how to convert String in hex representation to Int
As of Swift 2.0, every Int type has an initializer able to take String as an input. You can then easily generate an UnicodeScalar corresponding and print it afterwards. Without having to change your representation of chars as string ;).
UPDATED: Swift 3.0 changed UnicodeScalar initializer
print("\u{1f44d}") // 👍
let charAsString = "1f44d" // code in variable
let charAsInt = Int(charAsString, radix: 16)! // As indicated by #MartinR radix is required, default won't do it
let uScalar = UnicodeScalar(charAsInt)! // In Swift 3.0 this initializer is failible so you'll need either force unwrap or optionnal unwrapping
print("\(uScalar)")
You can use
let char = "-12"
print(char.unicodeScalars.map {$0.value }))
You'll get the values as:
[45, 49, 50]
Here are a couple ways to do it:
let string = "1f44d"
Solution 1:
"&#x\(string);".applyingTransform(.toXMLHex, reverse: true)
Solution 2:
"U+\(string)".applyingTransform(StringTransform("Hex/Unicode"), reverse: true)
I made this extension that works pretty well:
extension String {
var unicode: String? {
if let charCode = UInt32(self, radix: 16),
let unicode = UnicodeScalar(charCode) {
let str = String(unicode)
return str
}
return nil
}
}
How to test it:
if let test = "e9c8".unicode {
print(test)
}
//print:
You cannot use string interpolation in Swift as you try to use it. Therefore, the following code won't compile:
let charAsString = "1f44d"
print("\u{\(charAsString)}")
You will have to convert your string variable into an integer (using init(_:radix:) initializer) then create a Unicode scalar from this integer. The Swift 5 Playground sample code below shows how to proceed:
let validCodeString = "1f44d"
let validUnicodeScalarValue = Int(validCodeString, radix: 16)!
let validUnicodeScalar = Unicode.Scalar(validUnicodeScalarValue)!
print(validUnicodeScalar) // 👍
I am very much new to swift language. I am performing some business logic which needs to take NSRange from given String.
Here is my requirement,
Given Amount = "144.44"
Need NSRange of only cent part i.e. after "."
Is there any API available for doing this?
You can do a regex-based search to find the range:
let str : NSString = "123.45"
let rng : NSRange = str.range("(?<=[.])\\d*$", options: .RegularExpressionSearch)
Regular expression "(?<=[.])\\d*$" means "zero or more digits following a dot character '.' via look-behind, all the way to the end of the string $."
If you want a substring from a given string you can use componentsSeparatedByString
Example :
var number: String = "144.44";
var numberresult= number.componentsSeparatedByString(".")
then you can get components as :
var num1: String = numberresult [0]
var num2: String = numberresult [1]
hope it help !!
Use rangeOfString and substringFromIndex:
let string = "123.45"
if let index = string.rangeOfString(".") {
let cents = string.substringFromIndex(index.endIndex)
print("\(cents)")
}
Another version that uses Swift Ranges, rather than NSRange
Define the function that returns an optional Range:
func centsRangeFromString(str: String) -> Range<String.Index>? {
let characters = str.characters
guard let dotIndex = characters.indexOf(".") else { return nil }
return Range(dotIndex.successor() ..< characters.endIndex)
}
Which you can test with:
let r = centsRangeFromString(str)
// I don't recommend force unwrapping here, but this is just an example.
let cents = str.substringWithRange(r!)
As the title says, what is the correct way to convert UnsafeMutablePointer to String in swift?
//lets say x = UnsafeMutablePointer<Int8>
var str = x.memory.????
I tried using x.memory.description obviously it is wrong, giving me a wrong string value.
If the pointer points to a NUL-terminated C string of UTF-8 bytes, you can do this:
import Foundation
let x: UnsafeMutablePointer<Int8> = ...
// or UnsafePointer<Int8>
// or UnsafePointer<UInt8>
// or UnsafeMutablePointer<UInt8>
let str = String(cString: x)
Times have changed. In Swift 3+ you would do it like this:
If you want the utf-8 to be validated:
let str: String? = String(validatingUTF8: c_str)
If you want utf-8 errors to be converted to the unicode error symbol: �
let str: String = String(cString: c_str)
Assuming c_str is of type UnsafePointer<UInt8> or UnsafePointer<CChar> which is the same type and what most C functions return.
this:
let str: String? = String(validatingUTF8: c_str)
doesn't appear to work with UnsafeMutablePointer<UInt8>
(which is what appears to be in my data).
This is me trivially figuring out how to do something like the C/Perl system function:
let task = Process()
task.launchPath = "/bin/ls"
task.arguments = ["-lh"]
let pipe = Pipe()
task.standardOutput = pipe
task.launch()
let data = pipe.fileHandleForReading.readDataToEndOfFile()
var unsafePointer = UnsafeMutablePointer<Int8>.allocate(capacity: data.count)
data.copyBytes(to: unsafePointer, count: data.count)
let output : String = String(cString: unsafePointer)
print(output)
//let output : String? = String(validatingUTF8: unsafePointer)
//print(output!)
if I switch to validatingUTF8 (with optional) instead of cString, I get this error:
./ls.swift:19:37: error: cannot convert value of type 'UnsafeMutablePointer<UInt8>' to expected argument type 'UnsafePointer<CChar>' (aka 'UnsafePointer<Int8>')
let output : String? = String(validatingUTF8: unsafePointer)
^~~~~~~~~~~~~
Thoughts on how to validateUTF8 on the output of the pipe (so I don't get the unicode error symbol anywhere)?
(yes, I'm not doing proper checking of my optional for the print(), that's not the problem I'm currently solving ;-) ).
I am new to iOS. I am currently studying iOS using Objective-C and Swift.
To append a string in Objective-C I am using following code:
NSString *string1 = #"This is";
NSString *string2 = #"Swift Language";
NSString *appendString=[NSString stringWithFormat:#"%# %#",string1,string2];
NSLog(#"APPEND STRING:%#",appendString);
Anyone please guide me.
Its very simple:
For ObjC:
NSString *string1 = #"This is";
NSString *string2 = #"Swift Language";
ForSwift:
let string1 = "This is"
let string2 = "Swift Language"
For ObjC AppendString:
NSString *appendString=[NSString stringWithFormat:#"%# %#",string1,string2];
For Swift AppendString:
var appendString1 = "\(string1) \(string2)"
var appendString2 = string1+string2
Result:
print("APPEND STRING 1:\(appendString1)")
print("APPEND STRING 2:\(appendString2)")
Complete Code In Swift:
let string1 = "This is"
let string2 = "Swift Language"
var appendString = "\(string1) \(string2)"
var appendString1 = string1+string2
print("APPEND STRING1:\(appendString1)")
print("APPEND STRING2:\(appendString2)")
In Swift, appending strings is as easy as:
let stringA = "this is a string"
let stringB = "this is also a string"
let stringC = stringA + stringB
Or you can use string interpolation.
let stringC = "\(stringA) \(stringB)"
Notice there will now be whitespace between them.
Note: I see the other answers are using var a lot. The strings aren't changing and therefore should be declared using let. I know this is a small exercise, but it's good to get into the habit of best practices. Especially because that's a big feature of Swift.
let string2 = " there"
var instruction = "look over"
choice 1 :
instruction += string2;
println(instruction)
choice 2:
var Str = instruction + string2;
println(Str)
ref this
Add this extension somewhere:
extension String {
mutating func addString(str: String) {
self = self + str
}
}
Then you can call it like:
var str1 = "hi"
var str2 = " my name is"
str1.addString(str2)
println(str1) //hi my name is
A lot of good Swift extensions like this are in my repo here, check them out: https://github.com/goktugyil/EZSwiftExtensions
You can simply append string
like:
var worldArg = "world is good"
worldArg += " to live";
var string1 = "This is ";
var string2 = "Swift Language";
var appendString = string1 + string2;
println("APPEND STRING: \(appendString)");
According to Swift 4 Documentation,
String values can be added together (or concatenated) with the addition operator (+) to create a new String value:
let string1 = "hello"
let string2 = " there"
var welcome = string1 + string2
// welcome now equals "hello there"
You can also append a String value to an existing String variable with the addition assignment operator (+=):
var instruction = "look over"
instruction += string2
// instruction now equals "look over there"
You can append a Character value to a String variable with the String type’s append() method:
let exclamationMark: Character = "!"
welcome.append(exclamationMark)
// welcome now equals "hello there!"
> Swift2.x:
String("hello ").stringByAppendingString("world") // hello world
Strings concatenate in Swift language.
let string1 = "one"
let string2 = "two"
var concate = " (string1) (string2)"
playgroud output is "one two"
In the accepted answer PREMKUMAR there are a couple of errors in his Complete code in Swift answer. First print should read (appendString) and Second print should read (appendString1). Also, updated println deprecated in Swift 2.0
His
let string1 = "This is"
let string2 = "Swift Language"
var appendString = "\(string1) \(string2)"
var appendString1 = string1+string2
println("APPEND STRING1:\(appendString1)")
println("APPEND STRING2:\(appendString2)")
Corrected
let string1 = "This is"
let string2 = "Swift Language"
var appendString = "\(string1) \(string2)"
var appendString1 = string1+string2
print("APPEND STRING:\(appendString)")
print("APPEND STRING1:\(appendString1)")
SWIFT 2.x
let extendedURLString = urlString.stringByAppendingString("&requireslogin=true")
SWIFT 3.0
From Documentation:
"You can append a Character value to a String variable with the String type’s append() method:" so we cannot use append for Strings.
urlString += "&requireslogin=true"
"+" Operator works in both versions
let extendedURLString = urlString+"&requireslogin=true"
let firstname = "paresh"
let lastname = "hirpara"
let itsme = "\(firstname) \(lastname)"