I want to see if a string is empty, and if so, insert the number "1".
how do i do this in swift 3?
perhaps something like this:
var example = String()
if example.isEmpty {
example = "1"
}
For some further reading about the basic empty string syntax in swift i'd recommend: https://www.dotnetperls.com/isempty-swift
Related
//First way
var myVar: String = " Hello"
print(myVar)
//Second way
var str = "Hello"
print(str)
I get the same output no matter which of the two I use. What's the difference between them?
These two are basically the same.
When you use var myVar: String = "Hello", you directly tell the swift compiler that your variable is type String.
When you use var myVar = "Hello", you do not specify the type of your variable, so the swift compiler has do do that for you.
Usually, you can get away without declaring your variable type and just have swift do it for you. However, in some cases, namely computed properties and custom classes/structures, you must manually declare your variable to be a specific type.
In your case, either way is fine. The end result is the same, just be aware of the difference for the future.
These two variable are same but with different names :
var myVar
var str
in swift Type doesn't matter that defined because, swift based on value it recognizes what type it is.
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 have a basic array that I'd like to manipulate. It seems that within the brackets of my if statement I can successfully manipulate the array (the right side of the screen shows that value has changed); however, once I leave the if statement, the changes seem to go away (the right side of the screen shows the original value, not the changed one). Thank you for your help!
var originalInfo = """
This
Is
A
Test
String
"""
// split string
var SplitInfo = originalInfo.split(separator: "\n")
// Manipulate Array
if SplitInfo[1].hasPrefix("I") == true {
SplitInfo[1].replacingOccurrences(of: "s", with: "s not")
}
SplitInfo[1]
The method replaceOccurences(of:with:) returns a String with the new value, so you'll need to assign the new value to the variable
SplitInfo[1] = SplitInfo[1].replacingOccurrences(of: "s", with: "s not")
Only then will you see the change. If you click on the method and check the Quick Help section in the right pane of Xcode, you'll see
Range<String.Index>? = default) -> String
the -> String means it returns a String, so you'll need to assign that to something.
If you're trying to maintain String then divide your string like this
var SplitInfo = originalInfo.components(separatedBy: "\n")
This will return [String] instead of substrings.
Hope this helps.
Edit based on OP's comment *
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 am now writing a program involves class and dictionaries. I wonder how could I access a class's values inside a dictionary. For the code below how do I access the test1 value using the dictionary. I have tried using dict[1].test1but it doesn't work.
class test {
var tes1 = 1
}
var refer = test()
var dict = [1:refer]
There are a few problems with the line dict[1].test1:
Firstly, the subscript on a dictionary returns an optional type because there may not be a value for the key. Therefore you need to check a value exists for that key.
Secondly, in your class Test you've defined a variable tes1, but you're asking for test1 from your Dictionary. This was possibly just a type-o though.
To solve these problems you're code should look something like this:
if let referFromDictionary = dict[1] {
prinln(referFromDictionary.test1)
}
That's because the subscript returns an optional, so you have to unwrap it - and the most straightforward way is by using optional chaining:
dict[1]?.tes1
but you can also use optional binding:
if let test = dict[1] {
let value = test.tes1
}