Is it possible to use a String/Character literal within string interpolation in Swift?
The language reference says:
The expression you write inside parentheses within an interpolated string cannot contain an unescaped double quote (") ...
That's a little fuzzy to me, since it seems to deliberately leave the loophole of escaped double quotes.
If I try:
println( "Output: \(repeat("H",20))" );
func repeat( char:Character, times:Int ) -> String {
var output:String = "";
for index in 1...times {
output += char;
}
return output;
}
I get "Expected ',' separator".
Similarly, if I do the same thing, but escape the quotes, still no dice:
println( "Output: \(repeat(\"H\",20))" );
I suspect it's just not possible, and to be honest, no big deal -- I haven't found any example that I can't easily solve by doing a little work before the string interpolation, I guess I'm just looking for confirmation that it's just not possible.
It can be done starting in Swift 2.1:
http://www.russbishop.net/swift-2-1
Prior to that, it wasn't possible.
I hope it's:
let strfunc = repeat("H",20) as string
println( "Output: \(strfunc)" );
Related
I have a string JSON, but it has fancy curly quotes in it which makes NSJSONSerialization fail.
let str = "{“title”:\"this is a “test” example\"}"
try! JSONSerialization.jsonObject(with: str.data(using: .utf8)!) // Error
The quotes around title are curly double quotes and apparently JSONSerialization can not handle it and fails. A naive approach would be to simple replace all instances of the curly quote with a non-curly one. The problem with that approach is that it will change the curly quotes around test which shouldn't be changed! The quotes around title are OK to be changed but the ones around test should not.
What can I do to get around this issue?
To fix this, you talk to whoever created the string, which does not contain JSON at the moment, and convince them to create a string that does contain JSON.
For JSON the rule is: If your parser can't parse it, then it's broken and you don't touch it.
The problem isn't that JSONSerialization cannot handle it. The problem is that JSONSerialization absolutely must not under any circumstances handle it. Because it's not JSON.
If curly quotes are only used for the keys, this regex will do the job:
let str = "{“title”:\"this is a “test” example\"}"
let strFixed = str.replacingOccurrences(
of: #"“(.[^”]*)”:\"(.[^\"]*)\""#,
with: "\"$1\":\"$2\"",
options: .regularExpression
)
// It's strongly recommended to use try/catch instead of force unwrapping.
let json = try! JSONSerialization.jsonObject(with: strFixed.data(using: .utf8)!)
If we print json, we get the correct result:
{
title = "this is a \U201ctest\U201d example";
}
Explanation
“(.[^”]*)”:\"(.[^\"]*)\"
------------------------
“(.[^”]*)” match everything between curly braces,
except the closing curling brace character
: separator between keys and values
\"(.[^\"]*)\" match everything between double quotes,
except the double quote character
\"$1\":\"$2\"
-------------
\"$1\" place the first captured group between double quotes
: separator between keys and values
\"$2\" place the second captured group between double quotes
I wanted to replace the first character of a String and got it to work like this:
s.replaceSubrange(Range(NSMakeRange(0,1),in:s)!, with:".")
I wonder if there is a simpler method to achieve the same result?
[edit]
Get nth character of a string in Swift programming language doesn't provide a mutable substring. And it requires writing a String extension, which isn't really helping when trying to shorten code.
To replace the first character, you can do use String concatenation with dropFirst():
var s = "😃hello world!"
s = "." + s.dropFirst()
print(s)
Result:
.hello world!
Note: This will not crash if the String is empty; it will just create a String with the replacement character.
Strings work very differently in Swift than many other languages. In Swift, a character is not a single byte but instead a single visual element. This is very important when working with multibyte characters like emoji (see: Why are emoji characters like 👩👩👧👦 treated so strangely in Swift strings?)
If you really do want to set a single random byte of your string to an arbitrary value as you expanded on in the comments of your question, you'll need to drop out of the string abstraction and work with your data as a buffer. This is sort of gross in Swift thanks to various safety features but it's doable:
var input = "Hello, world!"
//access the byte buffer
var utf8Buffer = input.utf8CString
//replace the first byte with whatever random data we want
utf8Buffer[0] = 46 //ascii encoding of '.'
//now convert back to a Swift string
var output:String! = nil //buffer for holding our new target
utf8Buffer.withUnsafeBufferPointer { (ptr) in
//Load the byte buffer into a Swift string
output = String.init(cString: ptr.baseAddress!)
}
print(output!) //.ello, world!
I'm very new to Swift; I've spent the morning reading StackOverflow and trying many strategies, in vain, to accomplish the following:
I have a string, say "12345 is your number!"
I want to extract "12345" to a variable.
In Java, I'd do something like:
String myStr = "12345 is your number!";
return myStr.substring(0, myStr.indexOf(" "));
How do I do something similar in Swift? I don't want to hard-code any assumptions about what the ending index will be. It might be 5 characters in, it might not. I just want to take the substring of everything up to the first occurrence of " ", wherever that might be.
The closest I've gotten so far is:
var myMessage = "12345 is your number!"
myMessage.endIndex.advancedBy(myMessage.characters.count - myMessage.characters.indexOf(" "))
but it doesn't compile for reasons I don't fully yet grok("Binary operator '-' cannot be applied to operands of type Distance (aka 'Int') and 'String.CharacterView.Index?'")
Any help on this is appreciated. Thank you.
Something like this should work:
myMessage.substringToIndex(myMessage.characters.indexOf(" ")!)
Note that in this code I force unwrapped the optional. If you're not guaranteed to have that space in the string, it might make more sense to have the index in a optional binding.
With optional binding, it would look something like this:
if let index = myMessage.characters.indexOf(" ") {
let result = myMessage.substringToIndex(index)
}
You can use a regex, try this code:
var myMessage = "12345 is your number!"
if let match = myMessage.rangeOfString("-?\\d+", options: .RegularExpressionSearch) {
print(myMessage.substringWithRange(match)) // 12345
let myNumber = Int(myMessage.substringWithRange(match)) // Then you can initialize a new variable
}
The advantage is that this method extracts only the numbers wherever they are in the String
Hope this help ;)
With Swift 5 you can use:
myStr.prefix(upTo: myStr.firstIndex(of: " ") ?? myStr.startIndex)
You may need to cast it back to String (String(myStr.prefix(upTo: myStr.firstIndex(of: " ") ?? myStr.startIndex))) since it returns a Substring
I'm very new to Swift; I've spent the morning reading StackOverflow and trying many strategies, in vain, to accomplish the following:
I have a string, say "12345 is your number!"
I want to extract "12345" to a variable.
In Java, I'd do something like:
String myStr = "12345 is your number!";
return myStr.substring(0, myStr.indexOf(" "));
How do I do something similar in Swift? I don't want to hard-code any assumptions about what the ending index will be. It might be 5 characters in, it might not. I just want to take the substring of everything up to the first occurrence of " ", wherever that might be.
The closest I've gotten so far is:
var myMessage = "12345 is your number!"
myMessage.endIndex.advancedBy(myMessage.characters.count - myMessage.characters.indexOf(" "))
but it doesn't compile for reasons I don't fully yet grok("Binary operator '-' cannot be applied to operands of type Distance (aka 'Int') and 'String.CharacterView.Index?'")
Any help on this is appreciated. Thank you.
Something like this should work:
myMessage.substringToIndex(myMessage.characters.indexOf(" ")!)
Note that in this code I force unwrapped the optional. If you're not guaranteed to have that space in the string, it might make more sense to have the index in a optional binding.
With optional binding, it would look something like this:
if let index = myMessage.characters.indexOf(" ") {
let result = myMessage.substringToIndex(index)
}
You can use a regex, try this code:
var myMessage = "12345 is your number!"
if let match = myMessage.rangeOfString("-?\\d+", options: .RegularExpressionSearch) {
print(myMessage.substringWithRange(match)) // 12345
let myNumber = Int(myMessage.substringWithRange(match)) // Then you can initialize a new variable
}
The advantage is that this method extracts only the numbers wherever they are in the String
Hope this help ;)
With Swift 5 you can use:
myStr.prefix(upTo: myStr.firstIndex(of: " ") ?? myStr.startIndex)
You may need to cast it back to String (String(myStr.prefix(upTo: myStr.firstIndex(of: " ") ?? myStr.startIndex))) since it returns a Substring
I was working on a Swift tutorial and found that Swift has a strange way to handle multi-line statement.
First, I defined some extension to the standard String class:
extension String {
func replace(target: String, withString: String) -> String {
return self.stringByReplacingOccurrencesOfString(target, withString: withString)
}
func toLowercase() -> String {
return self.lowercaseString
}
}
This works as expected:
let str = "HELLO WORLD"
let s1 = str.lowercaseString.replace("hello", withString: "goodbye") // -> goodbye world
This doesn't work:
let s2 = str
.lowercaseString
.replace("hello", withString: "goodbye")
// Error: could not find member 'lowercaseString'
If I replace the reference to the lowercaseString property with a function call, it works again:
let s3 = str
.toLowercase()
.replace("hello", withString: "goodbye") // -> goodbye world
Is there anything in the Swift language specifications that prevent a property to be broken onto its own line?
Code at Swift Stub.
This is definitely a compiler bug. Issue has been resolved in Xcode 7 beta 3.
This feels like a compiler bug, but it relates to the fact that you can define prefix, infix, and postfix operators in Swift (but not the . operator, ironically enough). I don't know why it only gives you grief on the property and not the function call, but is a combination of two things:
the whitespace before and after the . (dot) operator for properties (only)
some nuance of this ever growing language that treats properties differently than function calls (even though functions are supposed to first class types).
I would file a bug to see what comes out of it, Swift is not supposed to by pythonic this way. That said, to work around it, you can either not break the property from the type, or you can add a white space before and after the . .
let s2 = str.lowercaseString
.replace("hello", withString: "goodbye")
let s3 = str
. lowercaseString
.replace("hello", withString: "goodbye")
Using semicolons is not mandatory in swift. And I think that the problems with multiline statements in swift are because of optional semicolons.
Note that swift does not support multiline strings. Check here: Swift - Split string over multiple lines
So maybe swift cannot handle multiline statements. I am not sure about this and this could be one of the reasons so I would appreciate if anyone else can help regarding this issue.