kotlin string concatenation no side effect - string

Why is string concatenation kept without side-effects?
fun main() {
var s1 = "abc"
val s2 = "def"
s1.plus(s2)
println(s1)
// s1 = s1.plus(s2)
// println(s1)
}
I expected abcdef, but got just abc. The commented code works fine, but seems awkward.

The plus() method (or the + operator, which is basically the same thing in kotlin) is returning the concatenated string. So by calling
s1.plus(s2)
//or
s1 + s2
you concatenate these strings but you throw away the result.

As this answer shows. Your plus operation returns the concatenated string independent from the given strings. This is because strings are immutable in Kotlin. An alternative is to use CharLists or buffers.
Note that immutability has some serious benefits. With immutablility it is no problem to give away your references to immutable objects because nobody can change the object. You need to change the reference in your model to hold a new value. This way it helps a lot to guarantee consistency and integrity in your model. Also note that immutable objects are threadsafe for that reason.

Related

What is the meaning of "Object [RegExp String Iterator] {}" in NodeJs

I've been trying use matchAll in node.js, but when I run the code and log the return value it only shows Object [RegExp String Iterator] {}.
Could you help me to understand why this is the case?
Be mindful of the types of things you are working with.
To quote the docs for String.prototype.matchAll:
The matchAll() method returns an iterator of all results matching a string against a regular expression, including capturing groups.
(Emphasis mine.)
So, you get an iterator. What is an iterator? Well, the docs say:
In JavaScript an iterator is an object which defines a sequence and potentially a return value upon its termination.
[...]
While it is easy to imagine that all iterators could be expressed as arrays, this is not true. Arrays must be allocated in their entirety, but iterators are consumed only as necessary. Because of this, iterators can express sequences of unlimited size, such as the range of integers between 0 and Infinity.
So, matchAll will only do the actual work of finding the next match when you ask for it, by asking for next value of the iterator, and the amount of state that has to be kept won't increase that much with a longer string because not all matches have to be remembered at once. That's the beauty of iterators (and their opposite part, generators).
This is also why you won't see all the results in your console when printing the iterator - otherwise, a matchAll on a very very large string would cause a long delay and high CPU usage when its return value is merely logged to the console, which wouldn't make sense.
You can use the result of matchAll in a for of loop, which will look for the next match every time the loop repeats:
for (const match of 'abcde'.matchAll(/./g)) {
console.log(match)
}
// Prints 5 times something like ['a', index: 0, input: 'abcde', groups: undefined ]
Or, if you are willing to forgo the benefit of on-demand matching, you can extract all values from the iterator at once and fit them into an array using either spread syntax or Array.from:
const allMatches = [...'abcde'.matchAll(/./g)]
// - or -
const allMatches = Array.from('abcde'.matchAll(/./g))
(In fact, the spread syntax is shown in the example at the very top of the matchAll docs too.)

Fastest, leanest way to append characters to form a string in Swift

I come from a C# background where System.String is immutable and string concatenation is relatively expensive (as it requires reallocating the string) we know to use the StringBuilder type instead as it preallocates a larger buffer where single characters (Char, a 16-bit value-type) and short strings can be concatenated cheaply without extra allocation.
I'm porting some C# code to Swift which reads from a bit-array ([Bool]) at sub-octet indexes with character lengths less than 8 bits (it's a very space-conscious file format).
My C# code does something like this:
StringBuilder sb = new StringBuilder( expectedCharacterCount );
int idxInBits = 0;
Boolean[] bits = ...;
for(int i = 0; i < someLength; i++) {
Char c = ReadNextCharacter( ref idxInBits, 6 ); // each character is 6 bits in this example
sb.Append( c );
}
In Swift, I assume NSMutableString is the equivalent of .NET's StringBuilder, and I found this QA about appending individual characters ( How to append a character to string in Swift? ) so in Swift I have this:
var buffer: NSMutableString
for i in 0..<charCount {
let charValue: Character = readNextCharacter( ... )
buffer.AppendWithFormat("%c", charValue)
}
return String(buffer)
But I don't know why it goes through a format-string first, that seems inefficient (reparsing the format-string on every iteration) and as my code is running on iOS devices I want to be very conservative with my program's CPU and memory usage.
As I was writing this, I learned my code should really be using UnicodeScalar instead of Character, problem is NSMutableString does not let you append a UnicodeScalar value, you have to use Swift's own mutable String type, so now my code looks like:
var buffer: String
for i in 0..<charCount {
let x: UnicodeScalar = readNextCharacter( ... )
buffer.append(x)
}
return buffer
I thought that String was immutable, but I noticed its append method returns Void.
I still feel uncomfortable doing this because I don't know how Swift's String type is implemented internally, and I don't see how I can preallocate a large buffer to avoid reallocations (assuming Swift's String uses a growing algorithm).
(This answer was written based on documentation and source code valid for Swift 2 and 3: possibly needs updates and amendments once Swift 4 arrives)
Since Swift is now open-source, we can actually have a look at the source code for Swift:s native String
swift/stdlib/public/core/String.swift
From the source above, we have following comment
/// Growth and Capacity
/// ===================
///
/// When a string's contiguous storage fills up, new storage must be
/// allocated and characters must be moved to the new storage.
/// `String` uses an exponential growth strategy that makes `append` a
/// constant time operation *when amortized over many invocations*.
Given the above, you shouldn't need to worry about the performance of appending characters in Swift (be it via append(_: Character), append(_: UniodeScalar) or appendContentsOf(_: String)), as reallocation of the contiguous storage for a certain String instance should not be very frequent w.r.t. number of single characters needed to be appended for this re-allocation to occur.
Also note that NSMutableString is not "purely native" Swift, but belong to the family of bridged Obj-C classes (accessible via Foundation).
A note to your comment
"I thought that String was immutable, but I noticed its append method returns Void."
String is just a (value) type, that may be used by mutable as well as immutable properties
var foo = "foo" // mutable
let bar = "bar" // immutable
/* (both the above inferred to be of type 'String') */
The mutating void-return instance methods append(_: Character) and append(_: UniodeScalar) are accessible to mutable as well as immutable String instances, but naturally using them with the latter will yield a compile time error
let chars : [Character] = ["b","a","r"]
foo.append(chars[0]) // "foob"
bar.append(chars[0]) // error: cannot use mutating member on immutable value ...

Pass by Reference in Haskell?

Coming from a C# background, I would say that the ref keyword is very useful in certain situations where changes to a method parameter are desired to directly influence the passed value for value types of for setting a parameter to null.
Also, the out keyword can come in handy when returning a multitude of various logically unconnected values.
My question is: is it possible to pass a parameter to a function by reference in Haskell? If not, what is the direct alternative (if any)?
There is no difference between "pass-by-value" and "pass-by-reference" in languages like Haskell and ML, because it's not possible to assign to a variable in these languages. It's not possible to have "changes to a method parameter" in the first place in influence any passed variable.
It depends on context. Without any context, no, you can't (at least not in the way you mean). With context, you may very well be able to do this if you want. In particular, if you're working in IO or ST, you can use IORef or STRef respectively, as well as mutable arrays, vectors, hash tables, weak hash tables (IO only, I believe), etc. A function can take one or more of these and produce an action that (when executed) will modify the contents of those references.
Another sort of context, StateT, gives the illusion of a mutable "state" value implemented purely. You can use a compound state and pass around lenses into it, simulating references for certain purposes.
My question is: is it possible to pass a parameter to a function by reference in Haskell? If not, what is the direct alternative (if any)?
No, values in Haskell are immutable (well, the do notation can create some illusion of mutability, but it all happens inside a function and is an entirely different topic). If you want to change the value, you will have to return the changed value and let the caller deal with it. For instance, see the random number generating function next that returns the value and the updated RNG.
Also, the out keyword can come in handy when returning a multitude of various logically unconnected values.
Consequently, you can't have out either. If you want to return several entirely disconnected values (at which point you should probably think why are disconnected values being returned from a single function), return a tuple.
No, it's not possible, because Haskell variables are immutable, therefore, the creators of Haskell must have reasoned there's no point of passing a reference that cannot be changed.
consider a Haskell variable:
let x = 37
In order to change this, we need to make a temporary variable, and then set the first variable to the temporary variable (with modifications).
let tripleX = x * 3
let x = tripleX
If Haskell had pass by reference, could we do this?
The answer is no.
Suppose we tried:
tripleVar :: Int -> IO()
tripleVar var = do
let times_3 = var * 3
let var = times_3
The problem with this code is the last line; Although we can imagine the variable being passed by reference, the new variable isn't.
In other words, we're introducing a new local variable with the same name;
Take a look again at the last line:
let var = times_3
Haskell doesn't know that we want to "change" a global variable; since we can't reassign it, we are creating a new variable with the same name on the local scope, thus not changing the reference. :-(
tripleVar :: Int -> IO()
tripleVar var = do
let tripleVar = var
let var = tripleVar * 3
return()
main = do
let x = 4
tripleVar x
print x -- 4 :(

Is it possible to mutate a string in Swift such that it can be proven to modify the original value?

Start with the claim that Swift makes that strings are 'mutable', Are Swift "mutable" strings really mutable, or are they just like Java strings?, but proceeding with a generally-accepted (and non-Swift) definition of mutability - ie. strictly value immutability, without consideration of bindings
Is it possible to actually mutate a String value such that this prints 'true'?
var str = "Mutate me!"
let a1 = (unsafeAddressOf(str))
// some 'mutating operation'
let a2 = (unsafeAddressOf(str))
print(a1 == a2)
I am not interested in 'structure types' or delayed copy semantics. This question is about if a string value can be modified, from within Swift - although it would be also be interesting to see if such could be mutated by other .. devious means. (If not it is hogwash to even bother discussing value vs reference types instead of pure mutability considerations.)
I am aware of "mutating" methods but disagree with those as 'mutating' the string value. In the example above, let should be substitutable in the final program. If let cannot be substituted; or the object can otherwise not be proven to be the same, then mutability has not been shown.

Best way to build object from delimited string (hopefully not looped case)

this question feels like it would have been asked already, but I've not found anything so here goes...
I have constructor which is handed a string which is delimited. From that string I need to populate an object's instance variables. I can easily split the string by the delimited to give me an array of strings. I know I can simply iterate through the array and set my instance variables using ifs or a switch/case statement based on the current array index - however that just feels a bit nasty. Pseudo code:
String[] tokens = <from generic string tokenizer>;
for (int i = 0;i < tokens.length;i++) {
switch(i) {
case(0): instanceVariableA = tokens[i];
case(1): instanceVarliableB = tokens[i];
...
}
}
Does anyone have any ideas of how I do this better/nicer?
For what it's worth, I'm working in Java, but I guess this is language independant.
Uhm... "nasty" is in the way the constructor handles the parameters. If you can't change that then your code snippet is as good as it may be.
You could get rid of the for loop, though...
instanceVariableA = tokens[0];
instanceVariableB = tokens[1];
and then introduce constants (for readibilty):
instanceVariableA = tokens[VARIABLE_A_INDEX];
instanceVariableB = tokens[VARIABLE_B_INDEX];
NOTE: if you could change the string parameter syntax you could introduce a simple parser and, with a little bit of reflection, handle this thing in a slightly more elegant way:
String inputString = "instanceVariableA=some_stuff|instanceVariableB=some other stuff";
String[] tokens = inputString.split("|");
for (String token : tokens)
{
String[] elements = token.split("=");
String propertyName = tokens[0];
String propertyValue = tokens[1];
invokeSetter(this, propertyName, propertyValue); // TODO write method
}
Could you not use a "for-each" loop to eliminate much of the clutter?
I really think the way you are doing it is fine, and Manrico makes a good suggestion about using constants as well.
Another method would be to create a HashMap with integer keys and string values where the key is the index and the value is the name of the property. You could then use a simple loop and some reflection to set the properties. The reflection part might make this a bit slow, but in another language (say, PHP for example) this would be much cleaner.
just an untested idea,
keep the original token...
String[] tokens = <from generic string tokenizer>;
then create
int instanceVariableA = 0;
int instanceVariableB = 1;
if you need to use it, then just
tokens[instanceVariableA];
hence no more loops, no more VARIABLE_A_INDEX...
maybe JSON might help?
Python-specific solution:
Let's say params = ["instanceVariableA", "instanceVariableB"]. Then:
self.__dict__.update(dict(zip(params, tokens)))
should work; that's roughly equivalent to
for k,v in zip(params, tokens):
setAttr(self, k, v)
depending on the presence/absence of accessors.
In a non-dynamic language, you could accomplish the same effect building a mapping from strings to references/accessors of some kind.
(Also beware that zip stops when either list runs out.)

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