Add string value every-time it is called.

Add string value every-time it is called.

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Add string value every-time it is called.
Tag : actionscript-3 , By : PatrickSimonHenk
Date : November 14 2020, 05:18 PM

it should still fix some issue When my function buildingBuilt is called, it is dispatching an event like so: , Try updating your code like so,
  var moneyDropper:int = 0;

  private function timerMoney(event):void {

     if(moneyDropper == 0)
         moneyDropper += 500;

  private function dropMoney(event):void {
      money.text = String( Number(money.text ) - moneyDropper);

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Increment string in Textview each time class is called

Tag : android , By : socurious
Date : March 29 2020, 07:55 AM
To fix this issue Of course you can! Just create a getter and a setter method within your class.
public class Foo() {
private String fooString;

  public void setFooString() {
    fooString = fooString + "my new string";

  public String getFooString() {
    return fooString;

What happens to a function called in C from the time being called to the time it returns?

Tag : c , By : ralph okochu
Date : March 29 2020, 07:55 AM
like below fixes the issue That's unspecified; it's up to the implementation. As pointed out by Keith Thompson, it doesn't even have to tell you how it works. :)
Some implementations will put all the arguments on the stack, some will use registers, and many use a mix (the first n arguments passed in registers, any more and they go on the stack).

add one each time function is called --- string + number

Tag : javascript , By : CookingCoder
Date : March 29 2020, 07:55 AM
To fix the issue you can do Each time I call this function. I want the output to increase by one. The problem is my variable is a string... , You need to convert the string into a number then:
Number(localStorage.getItem('howManyLinks')) + 1
+localStorage.getItem('howManyLinks') + 1

String value gets converted to Time object when a model method is called using delayed job

Tag : ruby-on-rails , By : abuiles
Date : March 29 2020, 07:55 AM
like below fixes the issue This happens due to the fact the string looks like a datetime to Ruby's default YAML loader. DelayedJob uses YAML for serialization, so this aspect manifests.
If you want to keep the string as string, stuff some padding chars to the beginning of it, like x or something. Demo follows.
ds: 2017-06-21 07:17:00
>> require "yaml"
>> YAML.load_file("1.yml")
=> {"ds"=>2017-06-21 11:17:00 +0400}
>> _["ds"].class
=> Time

In loop condition: why is strlen() called every time, but vector.size() gets called only once?

Tag : cpp , By : pttr
Date : March 29 2020, 07:55 AM
I wish this helpful for you You are misinterpreting the generated assembly. It's not that the vector's size is assumed unchanged, it's just that the call to v.size() is inlined. The vector's size is still computed anew every loop iteration and loaded into rax.
The const qualifier only prevents the functions f and g from modifying the object they refer to. It does not mean they can assume the object is truly immutable and cannot change in size. The compiler has to assume that non-inline functions could be changing the vector, and that includes printf. So the function has to re-compute the vector's size in order to be correct.
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