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
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.
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.
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.