I've reproduced this problem in a Swift playground but haven't solved it yet...
I'd like to print one of a range of characters in a UILabel. If I explicitly declare the character, it works:
// This works.
let value: String = "\u{f096}"
label.text = value // Displays the referenced character.
However, I want to construct the String. The code below appears to produce the same result as the line above, except that it doesn't. It just produces the String \u{f096} and not the character it references.
// This doesn't work
let n: Int = 0x95 + 1
print(String(n, radix: 16)) // Prints "96".
let value: String = "\\u{f0\(String(n, radix: 16))}"
label.text = value // Displays the String "\u{f096}".
I'm probably missing something simple. Any ideas?
How about stop using string conversion voodoo and use standard library type UnicodeScalar?
You can also create Unicode scalar values directly from their numeric representation.
let airplane = UnicodeScalar(9992)
print(airplane)
// Prints "✈︎"
UnicodeScalar.init there is actually returning optional value, so you must unwrap it.
If you need String just convert it via Character type to String.
let airplaneString: String = String(Character(airplane)) // Assuming that airplane here is unwrapped
Related
just a short question. In Swift it is possible to solve the following code:
var a: String;
a = "\(3*3)";
The arithmetic operation in the string will be solved. But i can´t figure out, why this following variation doesn´t work.
var a: String;
var b: String;
b = "3*3";
a = "\(b)";
In this case the arithmetic operation in var a will not be resolved. Any ideas why and how i can this get to work. Some things would be much more easier if this would work. Thanks for your answers.
In the second case, you are interpolating a string, not an arithmetic expression. In your example, it's a string you chose at compile time, but in general it might be a string from the user, or loaded from a file or over the web. In other words, at runtime b could contain some arbitrary string. The compiler isn't available at runtime to parse an arbitrary string as arithmetic.
If you want to evaluate an arbitrary string as an arithmetic formula at runtime, you can use NSExpression. Here's a very simple example:
let expn = NSExpression(format:"3+3")
println(expn.expressionValueWithObject(nil, context: nil))
// output: 6
You can also use a third-party library like DDMathParser.
Swift 4.2
let expn = "3+3"
print(expn.expressionValue(with: nil, context: nil))
But I also have a solution thats not the most effective way but could be used in some cases if your sure it's only "y+x" and not longer string.
var yNumber: Int!
var xNumber: Int!
let expn: String? = "3+3"
// Here we take to first value in the expn String.
if let firstNumber = expo?.prefix(1), let myInt = Int(firstNumber){
// This will print (Int : 3)
print("Int : \(myInt)")
// I set the value to yNumber
yNumber = myInt
}
// Here we take the last value in the expn string
if let lastNumber = optionalString?.suffix(1), let myInt = Int(lastNumber){
// This will print (Int : 3)
print("Int : \(myInt)")
// I set the value to xNumber
xNumber = myInt
}
// Now you can take the two numbers and add
print(yNumber + xNumber)
// will print (6)
I can't recommend this but it works in some cases
This won't be solved because this is not an arithmetic operation, this is a string:
"3*3"
the same as this
"String"
Everything you put in " it's a string.
The second example lets you construct a new String value from a mix of constants, variables, literals, and expressions:
"\(3*3)"
this is possible because of string interpolation \()
You inserted a string expression which swing convert and create expected result.
You can try to use evaluatePostfixNotationString method from that class.
The whole project is about recognizing math expression from camera image and calculating it after.
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 just joined a project that has a lot of existing code. The previous programmer was perhaps unfamiliar with Swift or began development in the early stages of the Swift language. They seemed to be using the if let statement in an odd way. They seemed to want to use the statement as a if is let. Before I edit the code I would like to know if there is any valid use for this:
// In JSON parser
if value is String, let string = value as? String {
document.createdBy = string
}
First checking if value is of type String seems redundant to me. Doesn't Swift check for this in the let string = value as? String portion of the statement?
QUESTION
Why would this need to be checked twice? Or would there be a reason for this?
You're correct, this is redundant. If value is not a string, then value as? String would return nil, and the conditional binding would fail.
To check the type, and not use the casted result:
if value is String {
// Do something that doesn't require `value` as a string
}
To check the type and use the result:
if let value = value as? String { // The new name can shadow the old name
document.createdBy = value
}
Doing both makes no sense.
let first = postalText.text?[(postalText.text?.startIndex)!]
let second = postalText.text?[(postalText.text?.index((postalText.text?.startIndex)!, offsetBy: 1))!]
let third = postalText.text?[(postalText.text?.index((postalText.text?.startIndex)!, offsetBy: 2))!]
I'm trying to capitalize the FIRST and THIRD character and then merge all 3 into a new string
but the .uppercase and .capitalized doesn't work .
Also how do i check that the SECOND character is a number ?
.uppercased and .capitalized only work for strings, what you show there are Characters. You can cast a Character as a String and make it capitalized.
let firstCapitalized = String(first!).capitalized
If you want to check if a Character is an int, you can also make it a String, and then check if casting the String as an Int is non-nil:
if Int("\(second!)") != nil {
print("Is Integer")
}
These cases all assume your first, second, and third are all non-nil, and force-unwraps them.
EDIT
I had some free time and was overlooking some old posts on SO, and I realized this answer I posted isn't using the best coding form. First off, force unwrapping anything is always a bad idea (it's a recipe for a crash in the future), so for the first part. Do something like this:
let firstCapitalized = String(first ?? "").capitalized
This at least gives you a back-out in case first == nil then you'll just be stuck with an empty string.
For the second part, I would use optional unwrapping instead of if Int("\(second!)") != nil. I would say the more proper method would be something like this:
if let second = second, let stringConvertedToInteger = Int("\(String(second))") {
print("\(stringConvertedToInteger) is an integer")
} else {
print("Either second is nil, or it cannot be converted to an integer")
}
This will optionally unwrap the character second, and if it has a value, convert it to an integer (should it be one, checked by optional unwrapping). This is the safest way to do it, and will keep you from experiencing any runtime errors.
just a short question. In Swift it is possible to solve the following code:
var a: String;
a = "\(3*3)";
The arithmetic operation in the string will be solved. But i can´t figure out, why this following variation doesn´t work.
var a: String;
var b: String;
b = "3*3";
a = "\(b)";
In this case the arithmetic operation in var a will not be resolved. Any ideas why and how i can this get to work. Some things would be much more easier if this would work. Thanks for your answers.
In the second case, you are interpolating a string, not an arithmetic expression. In your example, it's a string you chose at compile time, but in general it might be a string from the user, or loaded from a file or over the web. In other words, at runtime b could contain some arbitrary string. The compiler isn't available at runtime to parse an arbitrary string as arithmetic.
If you want to evaluate an arbitrary string as an arithmetic formula at runtime, you can use NSExpression. Here's a very simple example:
let expn = NSExpression(format:"3+3")
println(expn.expressionValueWithObject(nil, context: nil))
// output: 6
You can also use a third-party library like DDMathParser.
Swift 4.2
let expn = "3+3"
print(expn.expressionValue(with: nil, context: nil))
But I also have a solution thats not the most effective way but could be used in some cases if your sure it's only "y+x" and not longer string.
var yNumber: Int!
var xNumber: Int!
let expn: String? = "3+3"
// Here we take to first value in the expn String.
if let firstNumber = expo?.prefix(1), let myInt = Int(firstNumber){
// This will print (Int : 3)
print("Int : \(myInt)")
// I set the value to yNumber
yNumber = myInt
}
// Here we take the last value in the expn string
if let lastNumber = optionalString?.suffix(1), let myInt = Int(lastNumber){
// This will print (Int : 3)
print("Int : \(myInt)")
// I set the value to xNumber
xNumber = myInt
}
// Now you can take the two numbers and add
print(yNumber + xNumber)
// will print (6)
I can't recommend this but it works in some cases
This won't be solved because this is not an arithmetic operation, this is a string:
"3*3"
the same as this
"String"
Everything you put in " it's a string.
The second example lets you construct a new String value from a mix of constants, variables, literals, and expressions:
"\(3*3)"
this is possible because of string interpolation \()
You inserted a string expression which swing convert and create expected result.
You can try to use evaluatePostfixNotationString method from that class.
The whole project is about recognizing math expression from camera image and calculating it after.