var result = (dimensione-base)/2
dimensione & base are two "Double?" number, how can i solve this?
i want to calculate this, but he always gives me this error:"Binary operator - cannot be applied to two "double?" operands.
Double? is an optional type. That means a variable of this type can be nil. So before making an operation with them, you need to make sure these variables actually hold a value :
if let dimensione = dimensione,
let base = base
{
var result = (dimensione-base)/2
// Use result here
}else{
// Do something here when dimensione or base is nil.
}
You could also prefer to assign them a default value when they are nil (but in this case it seems less secure) :
var result = ((dimensione ?? 0)-(base ?? 0))/2
Here, if nil, 0 will be used instead.
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 am playing with Arrays in playground and I am bit confused. Here is code:
var players = ["tob", "cindy", "mindy"] //["tob", "cindy", "mindy"]
print(players.isEmpty) // False
var currentPlayer = players.first // "tob"
print(currentPlayer) // "Optional("tob")\n"
Why does it says "Optional"?
I found explanation: "The property first actually returns an optional, because if the array were empty, first would return nil."
But it is not empty. .isEmpty //false, So I am not understanding this.
Thanks for help in advance.
The correct way to think of Optional is that this may or may not have a value. What is the first element of an empty list? There is no such thing. It is not a value. We call that lack of a value nil or .None.
In Swift a variable must have a specific type. So your example:
let currentPlayer = players.first
What is the type of currentPlayer? It may be a String, or it may be nothing at all. It is a "maybe string" and in Swift that's called an Optional<String>. Whether players has elements or doesn't have elements doesn't change the type of currentPlayer.
If you want to do something if-and-only-if the variable has a value, then there are many ways. The simplest is if-let.
let players = ["tob", "cindy", "mindy"] //["tob", "cindy", "mindy"]
print(players.isEmpty) // False
if let currentPlayer = players.first {
print(currentPlayer)
}
This will print tob as you're expecting.
Another very common approach is the guard let
let players = ["tob", "cindy", "mindy"] //["tob", "cindy", "mindy"]
guard let currentPlayer = players.first else { return }
print(currentPlayer)
This lets you avoid nesting the rest of your function inside of curly braces, but otherwise is the same approach.
It is possible to convert an Optional into its underlying type using !, but this is very dangerous and should be avoided except where absolutely necessary. Tools like if-let and guard-let (and also Optional.map) are almost always preferred.
But the key here is to understand that all Swift variables have a single type, and sometimes that type is "maybe it has a value, maybe it doesn't."
If we look at the description of first, we will see that it always returns optional type:
public var first: Self.Generator.Element? { get }
I have an "if let" statement that is being executed, despite the "let" part being nil.
if let leftInc : Double? = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!]! {
println(leftInc)
let valueString : String = formatter.stringFromNumber(NSNumber(double: leftInc!))!
self.leftIncisorTextField?.text = valueString
self.btnLeftIncisor.associatedLabel?.text = valueString
}
// self.analysis.inputs is a Dictionary<String, Double?>
The inputs dictionary holds information entered by the user - either a number, or nil if they haven't entered anything in the matching field yet.
Under the previous version of Swift, the code was written as this:
if let leftInc : Double? = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!]?? {
and worked correctly.
I saw a similar question here, but in that instance the problem seemed to be the result of using Any?, which is not the case here.
Swift 2.2
In your if let you define another optional, that's why nil is a legitimate case. if let is intended mainly to extract (maybe) non optional value from an optional.
You might try:
if let leftInc : Double = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!].flatMap ({$0}) {
// leftInc is not an optional in this scope
...
}
Anyway I'd consider to not do it as a one liner but take advantage of guard case. Just in order to enhance readability. And avoid bang operator (!).
The if-let is for unwrapping optionals. You are allowing nil values by setting the type to an optional Double.
The if statement should be:
if let leftInc = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!] as? Double{
...
}
This will attempt to get an object out of inputs, if that fails it returns nil and skips it. If it does return something it will attempt to convert it to a Double. If that fails it skips the if statement as well.
if inputs is a dictionary like [Something:Double] then you don't need the last as? Double as indexing the dictionary will return a Double?
I recommend reading the swift book on optional chaining.
You could break it down further -
if let optionalDouble = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!], leftInc = optionalDouble {
....
}
as your dictionary has optional values - this way of writing it might make it clearer what's going on
if let k = dict["someKey"]{}, dict["someKey"] will be an object of type Any
this can bypass a nill
So do a typecast to get it correct like if let k = dict["someKey"] as! String {}
I'm getting errors when concatenating string:
let likeKey = "like-" + foodPhotoObjects[indexPath.row].objectId
Error
binary operator '+' cannot be applied to operands of type 'String' and 'String?!'
So, you have an implicitly-wrapped optional of an optional string, something like this:
struct Thing {
let objectId: String?!
}
let foodPhotoObjects: [Thing] = [Thing(objectId: "2")]
With any doubly-wrapped optional, to get to the object inside you’d need to unwrap it twice:
// first unwrap the String?! into a String?
if let outer = foodPhotoObjects[0].objectId,
// then unwrap that String? into a String
inner = outer {
// inner is now a String
println("like-\(inner)")
}
The key here is even though the outer optional is implicit (i.e. ! rather than ?), you can still unwrap implicit optionals using if let, so the implicitness is irrelevant when doing this.
An alternative way of handling this kind of thing, rather than if-let, is to use map:
let concatedString = foodPhotoObjects[indexPath.row].objectId.map {
"like-" + $0
} ?? ""
map on an optional means: if the optional contains a value, change the value using this function and return that as an optional, otherwise return nil. So, unwrap the String? and prepend “like” to it.
?? on an optional means: if the preceding value is nil, replace it with the default on the right-hand side (the empty string), otherwise unwrap it and return that (i.e. the value we just mapped).
Now for the tricky part: because the value we’re calling map on is an implicit optional, it will be implicitly unwrapped – that is, the map is being called on the inner String? rather than on the String?!. This is unlike the case with if let where that was run on the implicit optional first, then the inner optional.
As with all implicit optionals, there’s a risk that they might actually be nil in which case your code would blow up, like so:
let explode = Thing(objectId: nil)
// the next line will generate fatal error: unexpectedly
// found nil while unwrapping an Optional value
explode.objectId.map { "like-" + $0 }
If this is a concern, you could guard against it with some optional chaining:
// note, ? after objectId
let concatedString = foodPhotoObjects[indexPath.row].objectId?.map {
"like-" + $0
} ?? ""
This snippet could win a prize for most optional-handling techniques crammed into a single statement… but it should do what you need.
Swift does not do implicit conversion, even if both are of same type and one of them is of optional type.
Try this.
var concatedString = ""
if let foodphoto = foodPhotoObjects[indexPath.row].objectId as? String {
concatedString = "like-" + foodphoto
}
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.