Is String.substringWithRange in bounds? - swift

I'm having a problem with non-latin character sets, and I need to check that a Range is in bounds before performing .substringWithRange. This seems really easy, but I can't find the way to do this.
Given a range:
let lastCharRange = currentString.endIndex.predecessor() ..< currentString.endIndex
How can I check:
let lastExpected = expectedString.substringWithRange(lastCharRange)
is in bounds?

Indexes are tied closely to the String that generated them -- actually to the String's CharacterView, which is a CollectionType. This holds true generally for collections.
So, you simply can't use the Index you got from one String on another String.
Depending on what you are doing, you might have to get a substring from the first and then search the second. You can also get the two Strings' CharacterViews and work with them via their collection-based interface: expectedString.characters.last, for example.

Related

Trouble Understanding Xcodemath error messages

I have been working to understand why I am getting the error messages shown in the attachment.
The bottom-most message that indicates a comma is needed makes no sense to me at all.
The other two messages may well be related to a problem with data types, but I cannot determine what data type rules I have violated.
Many thanks for your time and attention.
It's a few different errors cropping up, and the error about the separator is not really indicative of the problem.
SecondPartFraction is being declared twice. If those are meant to be two different variables, they should have two different names. If you simply wish to reassign a new value to SecondPartFraction, just drop the var off the second time you use it (as is has already been declared, you simply need to refer to it again).
Doubles and Ints can't be together for division, so that error is correct. If you want to get a Double result, just change the 16 to 16.0. Then the compiler won't complain.
The numbers you're getting are originating from a textfield too, which might cause some problems. If the user enters text into your textfields, instead of numbers, the app will crash since StepFirstPart and StepSecondPart are force unwrapped. You will probably want to do some kind of Optional Chaining to handle the case where the entry is not numeric.
In the last line, the label text is being set to an Int - in order to do this, you'll have to use string interpolation instead, since the text for a label must be a string rather than a number:
TotalNumRisers.text = "\(TotalRisers)"
Just one last quick note - in Swift, camel casing is standard for naming, so the first letter of every variable should be lowercase, then the others upper. So StepFirstPart would become stepFirstPart instead.
You create the same variable twice here e.x
var x = 0
var x = value + x
instead it should be
var x = 0
x = value + x // remove var from here

Ambiguous use of 'lazy'

I have no idea why this example is ambiguous. (My apologies for not adding the code here, it's simply too long.)
I have added prefix (_ maxLength) as an overload to LazyDropWhileBidirectionalCollection. subscript(position) is defined on LazyPrefixCollection. Yet, the following code from the above example shouldn't be ambiguous, yet it is:
print([0, 1, 2].lazy.drop(while: {_ in false}).prefix(2)[0]) // Ambiguous use of 'lazy'
It is my understanding that an overload that's higher up in the protocol hierarchy will get used.
According to the compiler it can't choose between two types; namely LazyRandomAccessCollection and LazySequence. (Which doesn't make sense since subscript(position) is not a method of LazySequence.) LazyRandomAccessCollection would be the logical choice here.
If I remove the subscript, it works:
print(Array([0, 1, 2].lazy.drop(while: {_ in false}).prefix(2))) // [0, 1]
What could be the issue?
The trail here is just too complicated and ambiguous. You can see this by dropping elements. In particular, drop the last subscript:
let z = [0, 1, 2].lazy.drop(while: {_ in false}).prefix(2)
In this configuration, the compiler wants to type z as LazyPrefixCollection<LazyDropWhileBidirectionalCollection<[Int]>>. But that isn't indexable by integers. I know it feels like it should be, but it isn't provable by the current compiler. (see below) So your [0] fails. And backtracking isn't powerful enough to get back out of this crazy maze. There are just too many overloads with different return types, and the compiler doesn't know which one you want.
But this particular case is trivially fixed:
print([0, 1, 2].lazy.drop(while: {_ in false}).prefix(2).first!)
That said, I would absolutely avoid pushing the compiler this hard. This is all too clever for Swift today. In particular overloads that return different types are very often a bad idea in Swift. When they're simple, yes, you can get away with it. But when you start layering them on, the compiler doesn't have a strong enough proof engine to resolve it. (That said, if we studied this long enough, I'm betting it actually is ambiguous somehow, but the diagnostic is misleading. That's a very common situation when you get into overly-clever Swift.)
Now that you describe it (in the comments), the reasoning is straightforward.
LazyDropWhileCollection can't have an integer index. Index subscripting is required to be O(1). That's the meaning of the Index subscript versus other subscripts. (The Index subscript must also return the Element type or crash; it can't return an Element?. That's way there's a DictionaryIndex that's separate from Key.)
Since the collection is lazy and has an arbitrary number of missing elements, looking up any particular integer "count" (first, second, etc.) is O(n). It's not possible to know what the 100th element is without walking through at least 100 elements. To be a collection, its O(1) index has to be in a form that can only be created by having previously walked the sequence. It can't be Int.
This is important because when you write code like:
for i in 1...1000 { print(xs[i]) }
you expect that to be on the order of 1000 "steps," but if this collection had an integer index, it would be on the order of 1 million steps. By wrapping the index, they prevent you from writing that code in the first place.
This is especially important in highly generic languages like Swift where layers of general-purpose algorithms can easily cascade an unexpected O(n) operation into completely unworkable performance (by "unworkable" I mean things that you expected to take milliseconds taking minutes or more).
Change the last row to this:
let x = [0, 1, 2]
let lazyX: LazySequence = x.lazy
let lazyX2: LazyRandomAccessCollection = x.lazy
let lazyX3: LazyBidirectionalCollection = x.lazy
let lazyX4: LazyCollection = x.lazy
print(lazyX.drop(while: {_ in false}).prefix(2)[0])
You can notice that the array has 4 different lazy conformations - you will have to be explicit.

NSRange from first occurrence until end of string

I'm pulling my hair out trying to generate a valid NSRange, it doesn't seem like it should be this complicated so I'm guessing I'm using the wrong approach. Here is what I'm trying to do:
I have a string with some unicode character in it:
"The quick brown fox\n❄jumped\n❄over the lazy dog"
I want to create an NSRange from that character until the end of string, and while I can get the corresponding index for the first occurrence of the character:
text.rangeOfString("❄")?.startIndex
I can't seem to get the end of the string in a consistent format (something that I can pass to NSMakeRange) to actually generate the range. This seems like it should be pretty simple, yet I've been stuck for over an hour now trying to figure out how to get it to work, I keep ending up with Index types that I can't cast to integers to convert back to length that NSMakeRange requires for its second element.
Ideally I'd do something like this (which is invalid due to incompatible and non-castable types (Index vs Int)):
let start = text.rangeOfString("❄")?.startIndex
NSMakeRange(start, text.endIndex - start)
I am using Swift, so I have the ability to use Swift's Range<String.Index>, if it will make things easier, although it seems to be yet another range representation different from NSRange and I'm not sure how compatible the two are (don't want to run into another dimension of Index vs Int).
Cast your String as NSString.
You will be able to use Foundation's .rangeOfString instead of Swift's .rangeOfString.
The Foundation's one will return an NSRange.
Be careful though, it doesn't work the same as Swift's method with Unicode, and NSRange and Range are not compatible (although there's ways to convert them).

Swift: Converting a string into a variable name

I have variables with incremented numbers within, such as row0text, row1text, row2text, etc.
I've figured out how to dynamically create string versions of those variable names, but once I have those strings, how can I use them as actual variable names rather than strings in my code?
Example:
var row3text = "This is the value I need!"
var firstPart = "row"
var rowNumber = 3
var secondPart = "text"
var together = (firstPart+String(rowNumber)+secondPart)
// the below gives me the concatenated string of the three variables, but I'm looking for a way to have it return the value set at the top.
println (together)
Once I know how to do this, I'll be able to iterate through those variables using a for loop; it's just that at the moment I'm unsure of how to use that string as a variable name in my code.
Thanks!
Short Answer: There is no way to do this for good reason. Use arrays instead.
Long Answer:
Essentially you are looking for a way to define an unknown number of variables that are all linked together by their common format. You are looking to define an ordered set of elements of variable length. Why not just use an array?
Arrays are containers that allow you to store an ordered set or list of elements and access them by their ordered location, which is exactly what you're trying to do. See Apple's Swift Array Tutorial for further reading.
The advantage of arrays is that they are faster, far more convenient for larger sets of elements (and probably the same for smaller sets as well), and they come packaged with a ton of useful functionality. If you haven't worked with arrays before it is a bit of a learning curve but absolutely worth it.

How to delete elements from a transformed collection using a predicate?

If I have an ArrayList<Double> dblList and a Predicate<Double> IS_EVEN I am able to remove all even elements from dblList using:
Collections2.filter(dblList, IS_EVEN).clear()
if dblList however is a result of a transformation like
dblList = Lists.transform(intList, TO_DOUBLE)
this does not work any more as the transformed list is immutable :-)
Any solution?
Lists.transform() accepts a List and helpfully returns a result that is RandomAccess list. Iterables.transform() only accepts an Iterable, and the result is not RandomAccess. Finally, Iterables.removeIf (and as far as I see, this is the only one in Iterables) has an optimization in case that the given argument is RandomAccess, the point of which is to make the algorithm linear instead of quadratic, e.g. think what would happen if you had a big ArrayList (and not an ArrayDeque - that should be more popular) and kept removing elements from its start till its empty.
But the optimization depends not on iterator remove(), but on List.set(), which is cannot be possibly supported in a transformed list. If this were to be fixed, we would need another marker interface, to denote that "the optional set() actually works".
So the options you have are:
Call Iterables.removeIf() version, and run a quadratic algorithm (it won't matter if your list is small or you remove few elements)
Copy the List into another List that supports all optional operations, then call Iterables.removeIf().
The following approach should work, though I haven't tried it yet.
Collection<Double> dblCollection =
Collections.checkedCollection(dblList, Double.class);
Collections2.filter(dblCollection, IS_EVEN).clear();
The checkCollection() method generates a view of the list that doesn't implement List. [It would be cleaner, but more verbose, to create a ForwardingCollection instead.] Then Collections2.filter() won't call the unsupported set() method.
The library code could be made more robust. Iterables.removeIf() could generate a composed Predicate, as Michael D suggested, when passed a transformed list. However, we previously decided not to complicate the code by adding special-case logic of that sort.
Maybe:
Collection<Double> odds = Collections2.filter(dblList, Predicates.not(IS_EVEN));
or
dblList = Lists.newArrayList(Lists.transform(intList, TO_DOUBLE));
Collections2.filter(dblList, IS_EVEN).clear();
As long as you have no need for the intermediate collection, then you can just use Predicates.compose() to create a predicate that first transforms the item, then evaluates a predicate on the transformed item.
For example, suppose I have a List<Double> from which I want to remove all items where the Integer part is even. I already have a Function<Double,Integer> that gives me the Integer part, and a Predicate<Integer> that tells me if it is even.
I can use these to get a new predicate, INTEGER_PART_IS_EVEN
Predicate<Double> INTEGER_PART_IS_EVEN = Predicates.compose(IS_EVEN, DOUBLE_TO_INTEGER);
Collections2.filter(dblList, INTEGER_PART_IS_EVEN).clear();
After some tries, I think I've found it :)
final ArrayList<Integer> ints = Lists.newArrayList(1, 2, 3, 4, 5);
Iterables.removeIf(Iterables.transform(ints, intoDouble()), even());
System.out.println(ints);
[1,3,5]
I don't have a solution, instead I found some kind of a problem with Iterables.removeIf() in combination with Lists.TransformingRandomAccessList.
The transformed list implements RandomAccess, thus Iterables.removeIf() delegates to Iterables.removeIfFromRandomAccessList() which depends on an unsupported List.set() operation.
Calling Iterators.removeIf() however would be successful, as the remove() operation IS supported by Lists.TransformingRandomAccessList.
see: Iterables: 147
Conclusion: instanceof RandomAccess does not guarantee List.set().
Addition:
In special situations calling removeIfFromRandomAccessList() even works:
if and only if the elements to erase form a compact group at the tail of the List or all elements are covered by the Predicate.