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new array list created everytime function is called recursively


By : marlon tayag
Date : October 17 2020, 06:10 AM
hop of those help? Either pass a single ArrayList instance as an argument to your method (public static void strToList(String word, List letters) - your method doesn't have to return anything in this case), or merge the ArrayList instances as follows:
code :
public static ArrayList<Character> strToList(String word)
{
    ArrayList<Character> letters = new ArrayList<Character>();
    if (!word.isEmpty()) {
        char let = word.charAt(0);
        letters.add(Character.valueOf(let));
        System.out.println(letters);
        String temp = word.substring(1, word.length());
        System.out.println(temp); 
        letters.addAll(strToList(temp)); // add all the elements of the List returned by the
                                         // recursive call to the new List created in this 
                                         // call
    }
    return letters;
}


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Does this function create a new String everytime this is called?


By : Александр Павленко
Date : March 29 2020, 07:55 AM
will help you No, that won't create another string each time. String constants are interned - so actually the constant "NotSubscribed" will refer to the same string object throughout your code, not even just every time this method is executed.
From section 3.10.5 of the Java Language Specification:
code :
public List<String> createStrings() {
    List<String> list = new ArrayList<String>();
    for (int i = 0; i < 10; i++) {
        list.add("foo" + i);
    }
    return list;
}

The elegant solution for function that gives everytime called different element of array(ordered and starting from new w


By : user7333754
Date : March 29 2020, 07:55 AM
To fix this issue This process of "popping" is actually iteration... of a custom sequence. The appropriate way of representing this in Swift is as a type (struct/class) that implements the IteratorProtocol. I called mine CycleIterator. Iterators are rarely used directly. Rather, they're usually provided by a type that conforms to Sequence. I called mine CycleSequence
The Sequence protocol simply requires the conforming type to provide a function, makeIterator(), which returns an iterator (CycleIterator in my case). Simply by doing this, you instantly gain all of functionality of sequences. Iterability, map/filter/reduce, prefix, suffix, etc.
code :
public struct CycleSequence<C: Collection>: Sequence {
    public let cycledElements: C

    public init(cycling cycledElements: C) {
        self.cycledElements = cycledElements
    }

    public func makeIterator() -> CycleIterator<C> {
        return CycleIterator(cycling: cycledElements)
    }
}

public struct CycleIterator<C: Collection>: IteratorProtocol {
    public let cycledElements: C
    public private(set) var cycledElementIterator: C.Iterator

    public init(cycling cycledElements: C) {
        self.cycledElements = cycledElements
        self.cycledElementIterator = cycledElements.makeIterator()
    }

    public mutating func next() -> C.Iterator.Element? {
        if let next = cycledElementIterator.next() {
            return next
        } else {
            self.cycledElementIterator = cycledElements.makeIterator() // Cycle back again
            return cycledElementIterator.next()
        }
    }
}

let s1 = CycleSequence(cycling: [1, 2, 3]) // Works with arrays of numbers, as you would expect.
// Taking one element at a time, manually
var i1 = s1.makeIterator()
print(i1.next() as Any) // => Optional(1)
print(i1.next() as Any) // => Optional(2)
print(i1.next() as Any) // => Optional(3)
print(i1.next() as Any) // => Optional(1)
print(i1.next() as Any) // => Optional(2)
print(i1.next() as Any) // => Optional(3)
print(i1.next() as Any) // => Optional(1)

let s2 = CycleSequence(cycling: 2...5) // Works with any Collection. Ranges work!
// Taking the first 10 elements
print(Array(s2.prefix(10))) // => [2, 3, 4, 5, 2, 3, 4, 5, 2, 3]

let s3 = CycleSequence(cycling: "abc") // Strings are Collections, so those work, too!
s3.prefix(10).map{ "you can even map over me! \($0)" }.forEach{ print($0) }


print(Array(CycleSequence(cycling: [true, false]).prefix(7))) // => [true, false, true, false, true, false, true]
print(Array(CycleSequence(cycling: 1...3).prefix(7))) // => [1, 2, 3, 1, 2, 3, 1]
print(Array(CycleSequence(cycling: "ABC").prefix(7))) // => ["A", "B", "C", "A", "B", "C", "A"]
print(Array(CycleSequence(cycling: EmptyCollection<Int>()).prefix(7))) // => []
print(Array(zip(1...10, CycleSequence(cycling: "ABC")))) // => [(1, "A"), (2, "B"), (3, "C"), (4, "A"), (5, "B"), (6, "C"), (7, "A"), (8, "B"), (9, "C"), (10, "A")]
func makeCycleSequence<C: Collection>(for c: C) -> AnySequence<C.Iterator.Element> {
    return AnySequence(
        sequence(state: (elements: c, elementIterator: c.makeIterator()), next: { state in
            if let nextElement = state.elementIterator.next() {
                return nextElement
            }
            else {
                state.elementIterator = state.elements.makeIterator()
                return state.elementIterator.next()
            }
        })
    )
}

let repeater = makeCycleSequence(for: [1, 2, 3])
print(Array(repeater.prefix(10)))

Calling a function 1ms later everytime it's called?


By : Steven Box
Date : March 29 2020, 07:55 AM
I hope this helps you . I want to call a function in a setTimeout starting from 10 milliseconds to 11 milliseconds to 12 milliseconds and so on, using the variable in a for loop. It's supposed to create a thousand of the same element but with increasing time between each, but it creates them all at the same time. I put my code below. , I don’t know what your mistake was, but the following code works:
code :
var timeDelay=10;
function create() {
  
  // do something e.g. change HTML
  document.getElementById("output").innerHTML=timeDelay;
  
  if (timeDelay<1000) {
    timeDelay++;
    setTimeout(create, timeDelay);
  }
}
create();
<div id="output"></div>

If map/filter/reduce is used on data array for Flatlist, everytime when render is called all list item will be redraw?


By : Y.sh
Date : March 29 2020, 07:55 AM
this will help FlatList uses a key value from each data object (see keyExtractor property) to determine if the objects in the same order and a shallow equal to determine if render is needed.
See also the extraData property as a way of controlling render.

php - return two times everytime a function is called


By : Jamie
Date : March 29 2020, 07:55 AM
This might help you Don't use a global variable, use a static variable in the function in question:
code :
function count_calls() {
     static $count = 0;
     ++$count;
     return ($count % 2 === 0) ? $count : NULL ; 
}
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