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.
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.
Converting a String to Int returns an optional value but converting a Double to Int does not return an optional value. Why is that? I wanted to check if a double value is bigger than maximum Int value, but because converting function does not return an optional value, I am not be able to check by using optional binding.
var stringNumber: String = "555"
var intValue = Int(stringNumber) // returns optional(555)
var doubleNumber: Double = 555
var fromDoubleToInt = Int(doubleNumber) // returns 555
So if I try to convert a double number bigger than maximum Integer, it crashes instead of returning nil.
var doubleNumber: Double = 55555555555555555555
var fromDoubleToInt = Int(doubleNumber) // Crashes here
I know that there's another way to check if a double number is bigger than maximum Integer value, but I'm curious as why it's happening this way.
If we consider that for most doubles, a conversion to Int simply means dropping the decimal part:
let pieInt = Int(3.14159) // 3
Then the only case in which the Int(Double) constructor returns nil is in the case of an overflow.
With strings, converting to Int returns an optional, because generally, strings, such as "Hello world!" cannot be represented as an Int in a way that universally makes sense. So we return nil in the case that the string cannot be represented as an integer. This includes, by the way, values that can be perfectly represented as doubles or floats:
Consider:
let iPi = Int("3.14159")
let dPi = Double("3.14159")
In this case, iPi is nil while dPi is 3.14159. Why? Because "3.14159" doesn't have a valid Int representation.
But meanwhile, when we use the Int constructor which takes a Double and returns non-optional, we get a value.
So, if that constructor is changed to return an optional, why would it return 3 for 3.14159 instead of nil? 3.14159 can't be represented as an integer.
But if you want a method that returns an optional Int, returning nil when the Double would overflow, you can just write that method.
extension Double {
func toInt() -> Int? {
let minInt = Double(Int.min)
let maxInt = Double(Int.max)
guard case minInt ... maxInt = self else {
return nil
}
return Int(self)
}
}
let a = 3.14159.toInt() // returns 3
let b = 555555555555555555555.5.toInt() // returns nil
Failable initializers and methods with Optional return types are designed for scenarios where you, the programmer, can't know whether a parameter value will cause failure, or where verifying that an operation will succeed is equivalent to performing the operation:
let intFromString = Int(someString)
let valueFromDict = dict[someKey]
Parsing an integer from a string requires checking the string for numeric/non-numeric characters, so the check is the same as the work. Likewise, checking a dictionary for the existence of a key is the same as looking up the value for the key.
By contrast, certain operations are things where you, the programmer, need to verify upfront that your parameters or preconditions meet expectations:
let foo = someArray[index]
let bar = UInt32(someUInt64)
let baz: UInt = someUInt - anotherUInt
You can — and in most cases should — test at runtime whether index < someArray.count and someUInt64 < UInt32.max and someUInt > anotherUInt. These assumptions are fundamental to working with those kinds of types. On the one hand, you really want to design around them from the start. On the other, you don't want every bit of math you do to be peppered with Optional unwrapping — that's why we have types whose axioms are stated upfront.
I can't figure out why this 'optional' isn't working in scenario 1, but without the optional ? it works in scenario 2.
Using Swift v 1.2, xCode 6.2
var stuff = "6t"
// SCENARIO 1
// Why is this failing when stuff contains non-digit characters?
// i.e. it works if stuff = "45".
if let value: Int? = stuff.toInt() {
println("value \(value!)")
}
// SCENARIO 2
// This works!
if let value = stuff.toInt() {
println("val3 \(value)")
}
For Reference also see these SO Answers:
* I wonder if the Sift 1.2 example/Answer here is just plain wrong?
Swift - Converting String to Int
Converting String to Int in Swift
The first IF is always true.
Infact in both cases:
when the toInt() returns a valid Int
when returns nil
the if let will succeed (and yes, the first IF is useless).
Specifically in your code toInt() returns nil in both scenarios.
But in your first scenario you are simply accepting nil as a valid value to enter the THEN block.
There is no point of using if let value: Int?. If the if let works, then the value is an Int. There is no way that it could be nil. Therefore, you do not need to declare it as an optional.
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 want to pass a string to DDMathParser and store answer in another string variable using swift. I am totally new to iOS platform so I dont really know syntax of functions. Example:
var expr = "5+9*2" //Some Expression
var result = DDMathParser.evaluate(expr) // Result of Expression
Here is a simple example how to use DDMathParser from Swift:
let expr = "5+9*2" as NSString
let result = expr.numberByEvaluatingString().integerValue
println(result) // 23
If you need floating point results then replace integerValue
by doubleValue.