Why do th array range primitives consume their sources? - range

The range primitives that are dedicated to the built-in arrays consume their sources but one could easily design a range system that would rather be based on the .ptr of the source (at first look that's more flexible).
struct ArrayRange(T)
{
private T* _front;
private T* _back;
this(ref T[] stuff) {
_front = stuff.ptr;
_back = _front + stuff.length;
}
#property bool empty(){return _front == _back;}
#property T front(){ return *_front;}
#property T back(){return *_back;}
void popFront(){ ++_front;}
void popBack(){--_back;}
T[] array(){return _front[0 .. _back - _front];}
typeof(this) save() {
auto r = this.array.dup;
return typeof(this)(r);
}
}
void main(string[] args)
{
auto s = "bla".dup;
// here the source is 'fire-proofed'
auto rng = ArrayRange!char(s);
rng.popFront;
assert (s == "bla");
assert (rng.array == "la");
// default primitives: now the source is consumed
import std.array;
s.popFront;
assert (s == "la");
}
Why is the default system not based on pointer arithmetic since poping the front imply reallocations/less efficiency?
Any rationale for this design?

I agree with you, there is no reason to reallocate at each popFront. Good thing that's not what happens then!
The popFront mechanism is quite alike what you presented, and the source isn't consumed, only the array on which you call popFront (because, it is a pop after all). What you implemented is what happens when you slice an array: you get a reference range to the original array:
auto a = [1, 2, 3];
auto s = a[];
s.popFront;
assert(s == [2, 3]);
assert(a == [1, 2, 3]);
.dup is there to provide a copy of an array so that you can safely modify the copy without changing the original, so it copies the original array then provides an input range to this copy. Of course you can modify the copy (that's the point), and popFront will change it but still uses pointer arithmetic and without changing the source.
auto a = [1, 2, 3];
auto s = a.dup;
s.popFront;
assert(s = [2, 3]);
assert(a = [1, 2, 3]);
.dup may not seem very useful as such because we are using it with arrays, but it is really important when dealing with "pure" ranges as a range is often lazy not to consume it. As the copy of the range is consumed and not the initial one, we can safely pass this copy to a function and still have our initial lazy range intact.

Related

Why Iam getting ReferenceOutOfRangeException while PlayerPref a list in Unity [duplicate]

I have some code and when it executes, it throws a IndexOutOfRangeException, saying,
Index was outside the bounds of the array.
What does this mean, and what can I do about it?
Depending on classes used it can also be ArgumentOutOfRangeException
An exception of type 'System.ArgumentOutOfRangeException' occurred in mscorlib.dll but was not handled in user code Additional information: Index was out of range. Must be non-negative and less than the size of the collection.
What Is It?
This exception means that you're trying to access a collection item by index, using an invalid index. An index is invalid when it's lower than the collection's lower bound or greater than or equal to the number of elements it contains.
When It Is Thrown
Given an array declared as:
byte[] array = new byte[4];
You can access this array from 0 to 3, values outside this range will cause IndexOutOfRangeException to be thrown. Remember this when you create and access an array.
Array Length
In C#, usually, arrays are 0-based. It means that first element has index 0 and last element has index Length - 1 (where Length is total number of items in the array) so this code doesn't work:
array[array.Length] = 0;
Moreover please note that if you have a multidimensional array then you can't use Array.Length for both dimension, you have to use Array.GetLength():
int[,] data = new int[10, 5];
for (int i=0; i < data.GetLength(0); ++i) {
for (int j=0; j < data.GetLength(1); ++j) {
data[i, j] = 1;
}
}
Upper Bound Is Not Inclusive
In the following example we create a raw bidimensional array of Color. Each item represents a pixel, indices are from (0, 0) to (imageWidth - 1, imageHeight - 1).
Color[,] pixels = new Color[imageWidth, imageHeight];
for (int x = 0; x <= imageWidth; ++x) {
for (int y = 0; y <= imageHeight; ++y) {
pixels[x, y] = backgroundColor;
}
}
This code will then fail because array is 0-based and last (bottom-right) pixel in the image is pixels[imageWidth - 1, imageHeight - 1]:
pixels[imageWidth, imageHeight] = Color.Black;
In another scenario you may get ArgumentOutOfRangeException for this code (for example if you're using GetPixel method on a Bitmap class).
Arrays Do Not Grow
An array is fast. Very fast in linear search compared to every other collection. It is because items are contiguous in memory so memory address can be calculated (and increment is just an addition). No need to follow a node list, simple math! You pay this with a limitation: they can't grow, if you need more elements you need to reallocate that array (this may take a relatively long time if old items must be copied to a new block). You resize them with Array.Resize<T>(), this example adds a new entry to an existing array:
Array.Resize(ref array, array.Length + 1);
Don't forget that valid indices are from 0 to Length - 1. If you simply try to assign an item at Length you'll get IndexOutOfRangeException (this behavior may confuse you if you think they may increase with a syntax similar to Insert method of other collections).
Special Arrays With Custom Lower Bound
First item in arrays has always index 0. This is not always true because you can create an array with a custom lower bound:
var array = Array.CreateInstance(typeof(byte), new int[] { 4 }, new int[] { 1 });
In that example, array indices are valid from 1 to 4. Of course, upper bound cannot be changed.
Wrong Arguments
If you access an array using unvalidated arguments (from user input or from function user) you may get this error:
private static string[] RomanNumbers =
new string[] { "I", "II", "III", "IV", "V" };
public static string Romanize(int number)
{
return RomanNumbers[number];
}
Unexpected Results
This exception may be thrown for another reason too: by convention, many search functions will return -1 (nullables has been introduced with .NET 2.0 and anyway it's also a well-known convention in use from many years) if they didn't find anything. Let's imagine you have an array of objects comparable with a string. You may think to write this code:
// Items comparable with a string
Console.WriteLine("First item equals to 'Debug' is '{0}'.",
myArray[Array.IndexOf(myArray, "Debug")]);
// Arbitrary objects
Console.WriteLine("First item equals to 'Debug' is '{0}'.",
myArray[Array.FindIndex(myArray, x => x.Type == "Debug")]);
This will fail if no items in myArray will satisfy search condition because Array.IndexOf() will return -1 and then array access will throw.
Next example is a naive example to calculate occurrences of a given set of numbers (knowing maximum number and returning an array where item at index 0 represents number 0, items at index 1 represents number 1 and so on):
static int[] CountOccurences(int maximum, IEnumerable<int> numbers) {
int[] result = new int[maximum + 1]; // Includes 0
foreach (int number in numbers)
++result[number];
return result;
}
Of course, it's a pretty terrible implementation but what I want to show is that it'll fail for negative numbers and numbers above maximum.
How it applies to List<T>?
Same cases as array - range of valid indexes - 0 (List's indexes always start with 0) to list.Count - accessing elements outside of this range will cause the exception.
Note that List<T> throws ArgumentOutOfRangeException for the same cases where arrays use IndexOutOfRangeException.
Unlike arrays, List<T> starts empty - so trying to access items of just created list lead to this exception.
var list = new List<int>();
Common case is to populate list with indexing (similar to Dictionary<int, T>) will cause exception:
list[0] = 42; // exception
list.Add(42); // correct
IDataReader and Columns
Imagine you're trying to read data from a database with this code:
using (var connection = CreateConnection()) {
using (var command = connection.CreateCommand()) {
command.CommandText = "SELECT MyColumn1, MyColumn2 FROM MyTable";
using (var reader = command.ExecuteReader()) {
while (reader.Read()) {
ProcessData(reader.GetString(2)); // Throws!
}
}
}
}
GetString() will throw IndexOutOfRangeException because you're dataset has only two columns but you're trying to get a value from 3rd one (indices are always 0-based).
Please note that this behavior is shared with most IDataReader implementations (SqlDataReader, OleDbDataReader and so on).
You can get the same exception also if you use the IDataReader overload of the indexer operator that takes a column name and pass an invalid column name.
Suppose for example that you have retrieved a column named Column1 but then you try to retrieve the value of that field with
var data = dr["Colum1"]; // Missing the n in Column1.
This happens because the indexer operator is implemented trying to retrieve the index of a Colum1 field that doesn't exist. The GetOrdinal method will throw this exception when its internal helper code returns a -1 as the index of "Colum1".
Others
There is another (documented) case when this exception is thrown: if, in DataView, data column name being supplied to the DataViewSort property is not valid.
How to Avoid
In this example, let me assume, for simplicity, that arrays are always monodimensional and 0-based. If you want to be strict (or you're developing a library), you may need to replace 0 with GetLowerBound(0) and .Length with GetUpperBound(0) (of course if you have parameters of type System.Array, it doesn't apply for T[]). Please note that in this case, upper bound is inclusive then this code:
for (int i=0; i < array.Length; ++i) { }
Should be rewritten like this:
for (int i=array.GetLowerBound(0); i <= array.GetUpperBound(0); ++i) { }
Please note that this is not allowed (it'll throw InvalidCastException), that's why if your parameters are T[] you're safe about custom lower bound arrays:
void foo<T>(T[] array) { }
void test() {
// This will throw InvalidCastException, cannot convert Int32[] to Int32[*]
foo((int)Array.CreateInstance(typeof(int), new int[] { 1 }, new int[] { 1 }));
}
Validate Parameters
If index comes from a parameter you should always validate them (throwing appropriate ArgumentException or ArgumentOutOfRangeException). In the next example, wrong parameters may cause IndexOutOfRangeException, users of this function may expect this because they're passing an array but it's not always so obvious. I'd suggest to always validate parameters for public functions:
static void SetRange<T>(T[] array, int from, int length, Func<i, T> function)
{
if (from < 0 || from>= array.Length)
throw new ArgumentOutOfRangeException("from");
if (length < 0)
throw new ArgumentOutOfRangeException("length");
if (from + length > array.Length)
throw new ArgumentException("...");
for (int i=from; i < from + length; ++i)
array[i] = function(i);
}
If function is private you may simply replace if logic with Debug.Assert():
Debug.Assert(from >= 0 && from < array.Length);
Check Object State
Array index may not come directly from a parameter. It may be part of object state. In general is always a good practice to validate object state (by itself and with function parameters, if needed). You can use Debug.Assert(), throw a proper exception (more descriptive about the problem) or handle that like in this example:
class Table {
public int SelectedIndex { get; set; }
public Row[] Rows { get; set; }
public Row SelectedRow {
get {
if (Rows == null)
throw new InvalidOperationException("...");
// No or wrong selection, here we just return null for
// this case (it may be the reason we use this property
// instead of direct access)
if (SelectedIndex < 0 || SelectedIndex >= Rows.Length)
return null;
return Rows[SelectedIndex];
}
}
Validate Return Values
In one of previous examples we directly used Array.IndexOf() return value. If we know it may fail then it's better to handle that case:
int index = myArray[Array.IndexOf(myArray, "Debug");
if (index != -1) { } else { }
How to Debug
In my opinion, most of the questions, here on SO, about this error can be simply avoided. The time you spend to write a proper question (with a small working example and a small explanation) could easily much more than the time you'll need to debug your code. First of all, read this Eric Lippert's blog post about debugging of small programs, I won't repeat his words here but it's absolutely a must read.
You have source code, you have exception message with a stack trace. Go there, pick right line number and you'll see:
array[index] = newValue;
You found your error, check how index increases. Is it right? Check how array is allocated, is coherent with how index increases? Is it right according to your specifications? If you answer yes to all these questions, then you'll find good help here on StackOverflow but please first check for that by yourself. You'll save your own time!
A good start point is to always use assertions and to validate inputs. You may even want to use code contracts. When something went wrong and you can't figure out what happens with a quick look at your code then you have to resort to an old friend: debugger. Just run your application in debug inside Visual Studio (or your favorite IDE), you'll see exactly which line throws this exception, which array is involved and which index you're trying to use. Really, 99% of the times you'll solve it by yourself in a few minutes.
If this happens in production then you'd better to add assertions in incriminated code, probably we won't see in your code what you can't see by yourself (but you can always bet).
The VB.NET side of the story
Everything that we have said in the C# answer is valid for VB.NET with the obvious syntax differences but there is an important point to consider when you deal with VB.NET arrays.
In VB.NET, arrays are declared setting the maximum valid index value for the array. It is not the count of the elements that we want to store in the array.
' declares an array with space for 5 integer
' 4 is the maximum valid index starting from 0 to 4
Dim myArray(4) as Integer
So this loop will fill the array with 5 integers without causing any IndexOutOfRangeException
For i As Integer = 0 To 4
myArray(i) = i
Next
The VB.NET rule
This exception means that you're trying to access a collection item by index, using an invalid index. An index is invalid when it's lower than the collection's lower bound or greater than equal to the number of elements it contains. the maximum allowed index defined in the array declaration
Simple explanation about what a Index out of bound exception is:
Just think one train is there its compartments are D1,D2,D3.
One passenger came to enter the train and he have the ticket for D4.
now what will happen. the passenger want to enter a compartment that does not exist so obviously problem will arise.
Same scenario: whenever we try to access an array list, etc. we can only access the existing indexes in the array. array[0] and array[1] are existing. If we try to access array[3], it's not there actually, so an index out of bound exception will arise.
To easily understand the problem, imagine we wrote this code:
static void Main(string[] args)
{
string[] test = new string[3];
test[0]= "hello1";
test[1]= "hello2";
test[2]= "hello3";
for (int i = 0; i <= 3; i++)
{
Console.WriteLine(test[i].ToString());
}
}
Result will be:
hello1
hello2
hello3
Unhandled Exception: System.IndexOutOfRangeException: Index was outside the bounds of the array.
Size of array is 3 (indices 0, 1 and 2), but the for-loop loops 4 times (0, 1, 2 and 3). So when it tries to access outside the bounds with (3) it throws the exception.
A side from the very long complete accepted answer there is an important point to make about IndexOutOfRangeException compared with many other exception types, and that is:
Often there is complex program state that maybe difficult to have control over at a particular point in code e.g a DB connection goes down so data for an input cannot be retrieved etc... This kind of issue often results in an Exception of some kind that has to bubble up to a higher level because where it occurs has no way of dealing with it at that point.
IndexOutOfRangeException is generally different in that it in most cases it is pretty trivial to check for at the point where the exception is being raised. Generally this kind of exception get thrown by some code that could very easily deal with the issue at the place it is occurring - just by checking the actual length of the array. You don't want to 'fix' this by handling this exception higher up - but instead by ensuring its not thrown in the first instance - which in most cases is easy to do by checking the array length.
Another way of putting this is that other exceptions can arise due to genuine lack of control over input or program state BUT IndexOutOfRangeException more often than not is simply just pilot (programmer) error.
These two exceptions are common in various programming languages and as others said it's when you access an element with an index greater than the size of the array. For example:
var array = [1,2,3];
/* var lastElement = array[3] this will throw an exception, because indices
start from zero, length of the array is 3, but its last index is 2. */
The main reason behind this is compilers usually don't check this stuff, hence they will only express themselves at runtime.
Similar to this:
Why don't modern compilers catch attempts to make out-of-bounds access to arrays?

Data ranged subscribe strange behavior

I was playing with swift's Data in the following a small code:
var d = Data(count: 10)
d[5] = 3
let d2 = d[5..<8]
print("\(d2[0])")
To my surprise, this code throws exception on print() while the following code does not:
var d = Data(count: 10)
d[5] = 3
let d2 = d.subdata(in: 5..<8)
print("\(d2[0])")
I somehow understand why this happens, but I don't get why this is designed like this. When I use subdata() I get a whole copy of range, so indexing is valid from 0. But when I use range subscribe [], I get access to the requested range while indexing is the same as before. So in my first example d2[5] is 3.
But I wonder why it is designed like this? I don't want to make a copy of my data by using subdata() method. I just wanted to access a portion of my data with better indexing.
This is especially creates unexpected behaviors if you pass it to a function. For example, following code creates unexpected results and exceptions and you may not find out easily why:
func testit(idata: Data) {
if idata.count > 0 {
print("\(idata.count)")
print("\(idata[0])")
}
}
//...
var d = Data(count: 10)
d[5] = 3
let d2 = d[5..<8]
testit(idata: d2)
This code is really strange. Because if you debug your code, you see that print("\(idata.count)") prints 3 as size of idata which is correct, but accessing it with idata[0] creates exception.
Is there any reason for this design? I was expecting that I could access resulting Data from subscribe starting index 0 while it is not true. Can I do this without using subdata() which creates copy of data or using additional arguments to pass base of data slice?
d[5..<8] returns Data.Slice – which happens to be Data. Generally, slices share the indices with their base collection, as documented in Slice.
One possible reason for this design decision is that it guarantees that subscripting a slice is a O(1) operation (adding an offset for accessing the base collection is not necessarily O(1), e.g. not for strings.)
It is also convenient, as in this example to locate the text after the second occurrence of a character in a string:
let string = "abcdefgabcdefg"
// Find first occurrence of "d":
if let r1 = string.range(of: "d") {
// Find second occurrence of "d":
if let r2 = string[r1.upperBound...].range(of: "d") {
print(string[r2.upperBound...]) // efg
}
}
As a consequence, you must never assume that the indices of a collection are zero-based (unless documented, as for Array.startIndex). Use startIndex to get the first index, or first to get the first element.

Swift adding to end of array

I know that Array has the append(_:) method, but this mutates the original array.
Why doesn't Array also implement appending(_:) which would return a new array with that element appended? I've implemented this a few times now, but am wondering if there is a reason why it doesn't already exist?
(My only guess would be that this is to do with efficiency - if you used this method in a loop you would be copying your array multiple times?)
You can make an extension of Array do to this :
extension Array{
func appending(_ element:Element) -> Array<Element>{
var result = self
result.append(element)
return result
}
}
What are we discussing?
Starting from 'append' documentation:
/// Append `newElement` to the Array.
///
/// - Complexity: Amortized O(1) unless `self`'s storage is shared with another live array; O(`count`) if `self` does not wrap a bridged
NSArray; otherwise the efficiency is unspecified..
public mutating func append(newElement: Element)
You need to have knowledge about pointer and smart pointer.
M elements
var a = [1, 2, 3]
Is a simple array.
If you append N elements, complexity will be O(N)
func addElements() {
for elment in [4, 5, 6] {
a.append(element)
}
}
If you create a new array starting from a:
var b = a
At run time b points to a, so if you execute again addElements complexity will be O(M) in order to copy original array plus O(N) in order to add N elements.
Why documentation says?
"otherwise the efficiency is unspecified.."
Because in this example we have a simple array of integer, but the behaviour of copying array of custom objects is unpredictable.

Does Swift copy on mutation in this scenario?

Essentially what I want is a temporary alias for a class property to improve readability.
I'm in the situation described by the following code and I can't see a straightforward solution. What I want to avoid is y being copied on mutation and then copied back.
Renaming y would reduce the readability of the actual algorithm a lot.
Is the Swift compiler smart enough to not actually allocate new memory and how would I be able to know that?
If not, how to prevent copying?
class myClass {
var propertyWithLongDescriptiveName: [Float]
func foo() {
var y = propertyWithLongDescriptiveName
// mutate y with formulas where y corresponds to a `y` from some paper
// ...
propertyWithLongDescriptiveName = y
}
// ...
}
struct Array is a value types in Swift, which means that they are always
copied when assigned to another variable. However, each struct Array
contains pointers (not visible in the public interface) to the actual
element storage. Therefore after
var a = [1, 2, 3, 4]
var b = a
both a and b are (formally independent) values, but with pointers to the same element storage.
Only when one of them is mutated, a copy of the element storage is made.
This is called "copy on write" and for example explained in
Friday Q&A 2015-04-17: Let's Build Swift.Array
So after
b[0] = 17
a and b are values with pointers to different (independent) element storage.
Further mutation of b does not copy the element storage again
(unless b is copied to another variable). Finally, if you assign
the value back
a = b
the old element storage of a is released, and both values are pointers to the same storage again.
Therefore in your example:
var y = propertyWithLongDescriptiveName
// ... mutate y ...
propertyWithLongDescriptiveName = y
a copy of the element storage is made exactly once (assuming
that you don't copy y to an additional variable).
If the array size does not change then a possible approach could be
var propertyWithLongDescriptiveName = [1.0, 2.0, 3.0, 4.0]
propertyWithLongDescriptiveName.withUnsafeMutableBufferPointer { y in
// ... mutate y ...
y[0] = 13
}
print(propertyWithLongDescriptiveName) // [13.0, 2.0, 3.0, 4.0]
withUnsafeMutableBufferPointer() calls the closure with an
UnsafeMutableBufferPointer to the element storage.
A UnsafeMutableBufferPointer is a RandomAccessCollection and
therefore offers an array-like interface.
No, the Swift compiler is not that smart. All you need is a small test to see what it does:
class MyClass {
var propertyWithLongDescriptiveName: [Float] = [1,2]
func foo() {
var y = propertyWithLongDescriptiveName
y[0] = 3 // copied an mutated
print(y) // [3,2]
print(propertyWithLongDescriptiveName) // [1,2]
}
}
let mc = MyClass()
mc.foo()
You have 2 optons:
Change propertyWithLongDescriptiveName to NSMutableArray, which is a reference type
Accept the overhead cost of copy-and-mutate to trade for readability of your algorithm. In many cases memory allocation cost is minimal compared to your algorithm's.

How exactly does the "let" keyword work in Swift?

I've read this simple explanation in the guide:
The value of a constant doesn’t need to be known at compile time, but you must assign it a value exactly once.
But I want a little more detail than this. If the constant references an object, can I still modify its properties? If it references a collection, can I add or remove elements from it? I come from a C# background; is it similar to how readonly works (apart from being able to use it in method bodies), and if it's not, how is it different?
let is a little bit like a const pointer in C. If you reference an object with a let, you can change the object's properties or call methods on it, but you cannot assign a different object to that identifier.
let also has implications for collections and non-object types. If you reference a struct with a let, you cannot change its properties or call any of its mutating func methods.
Using let/var with collections works much like mutable/immutable Foundation collections: If you assign an array to a let, you can't change its contents. If you reference a dictionary with let, you can't add/remove key/value pairs or assign a new value for a key — it's truly immutable. If you want to assign to subscripts in, append to, or otherwise mutate an array or dictionary, you must declare it with var.
(Prior to Xcode 6 beta 3, Swift arrays had a weird mix of value and reference semantics, and were partially mutable when assigned to a let -- that's gone now.)
It's best to think of let in terms of Static Single Assignment (SSA) -- every SSA variable is assigned to exactly once. In functional languages like lisp you don't (normally) use an assignment operator -- names are bound to a value exactly once. For example, the names y and z below are bound to a value exactly once (per invocation):
func pow(x: Float, n : Int) -> Float {
if n == 0 {return 1}
if n == 1 {return x}
let y = pow(x, n/2)
let z = y*y
if n & 1 == 0 {
return z
}
return z*x
}
This lends itself to more correct code since it enforces invariance and is side-effect free.
Here is how an imperative-style programmer might compute the first 6 powers of 5:
var powersOfFive = Int[]()
for n in [1, 2, 3, 4, 5, 6] {
var n2 = n*n
powersOfFive += n2*n2*n
}
Obviously n2 is is a loop invariant so we could use let instead:
var powersOfFive = Int[]()
for n in [1, 2, 3, 4, 5, 6] {
let n2 = n*n
powersOfFive += n2*n2*n
}
But a truly functional programmer would avoid all the side-effects and mutations:
let powersOfFive = [1, 2, 3, 4, 5, 6].map(
{(num: Int) -> Int in
let num2 = num*num
return num2*num2*num})
Let
Swift uses two basic techniques to store values for a programmer to access by using a name: let and var. Use let if you're never going to change the value associated with that name. Use var if you expect for that name to refer to a changing set of values.
let a = 5 // This is now a constant. "a" can never be changed.
var b = 2 // This is now a variable. Change "b" when you like.
The value that a constant refers to can never be changed, however the thing that a constant refers to can change if it is an instance of a class.
let a = 5
let b = someClass()
a = 6 // Nope.
b = someOtherClass() // Nope.
b.setCookies( newNumberOfCookies: 5 ) // Ok, sure.
Let and Collections
When you assign an array to a constant, elements can no longer be added or removed from that array. However, the value of any of that array's elements may still be changed.
let a = [1, 2, 3]
a.append(4) // This is NOT OK. You may not add a new value.
a[0] = 0 // This is OK. You can change an existing value.
A dictionary assigned to a constant can not be changed in any way.
let a = [1: "Awesome", 2: "Not Awesome"]
a[3] = "Bogus" // This is NOT OK. You may not add new key:value pairs.
a[1] = "Totally Awesome" // This is NOT OK. You may not change a value.
That is my understanding of this topic. Please correct me where needed. Excuse me if the question is already answered, I am doing this in part to help myself learn.
First of all, "The let keyword defines a constant" is confusing for beginners who are coming from C# background (like me). After reading many Stack Overflow answers, I came to the conclusion that
Actually, in swift there is no concept of constant
A constant is an expression that is resolved at compilation time. For both C# and Java, constants must be assigned during declaration:
public const double pi = 3.1416; // C#
public static final double pi = 3.1416 // Java
Apple doc ( defining constant using "let" ):
The value of a constant doesn’t need to be known at compile time, but you must assign the value exactly once.
In C# terms, you can think of "let" as "readonly" variable
Swift "let" == C# "readonly"
F# users will feel right at home with Swift's let keyword. :)
In C# terms, you can think of "let" as "readonly var", if that construct was allowed, i.e.: an identifier that can only be bound at the point of declaration.
Swift properties:
Swift Properties official documentation
In its simplest form, a stored property is a constant or variable that is stored as part of an instance of a particular class or structure. Stored properties can be either variable stored properties (introduced by the varkeyword) or constant stored properties (introduced by the let keyword).
Example:
The example below defines a structure called FixedLengthRange, which describes a range of integers whose range length cannot be changed once it is created:
struct FixedLengthRange {
var firstValue: Int
let length: Int
}
Instances of FixedLengthRange have a variable stored property called firstValue and a constant stored property called length. In the example above, length is initialized when the new range is created and cannot be changed thereafter, because it is a constant property.