Static value in for loop

Static value in for loop

By : Ariana De Andrade
Date : October 28 2020, 08:00 PM
should help you out No because you used a static variable. A static variable is common to all the instances (or objects) of the class because it is a class level variable. In other words you can say that only a single copy of static variable is created and shared among all the instances of the class. Memory allocation for such variables only happens once when the class is loaded in the memory. https://beginnersbook.com/2013/05/static-variable/
when you create an object it holds the value "" When you call run str will hold ten as you can see the value printed. After the first iteration as str doesn't hold "" it will change its value to five.
code :

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Possible to loop "static" data in mysql loop

Possible to loop "static" data in mysql loop

By : user3090705
Date : March 29 2020, 07:55 AM
help you fix your problem As you can see below I am pulling information from a database and writting it to a text file in a structured format ( to everyone who has helped me out). , You can have MySQL write the data directly to a file using
code :
SELECT cards.card_id
FROM cards
INNER JOIN card_cheapest ON (cards.card_id = card_cheapest.card_id) 
ORDER BY card_id 
Static Newbie: Are non-static variables in a static class/method implied static?

Static Newbie: Are non-static variables in a static class/method implied static?

By : user3897611
Date : March 29 2020, 07:55 AM
will help you You're showing local variables. So no, those aren't static variables. Each time you invoke the method (including if it invoked itself recursively), you get a new set of variables. Different threads will not be sharing those variables. Note that this has nothing to do with the class being a static class. You need to differentiate between:
Local variables, which belong to the method that declares them Instance variables, associated with a particular instance of the declaring type Static variables, associated with the declaring type itself (and not with any specific instance)
Use static keyword in a for loop?

Use static keyword in a for loop?

By : Ron Desmarais
Date : March 29 2020, 07:55 AM
it fixes the issue I don't know why you want to use static keyword in your code, but as suggested you should read here: http://php.net/manual/en/language.oop5.static.php, how and when to use static keyword.
If your goal is to display dropdown and getting option value different 15mint time like this 12:00, 12:15, 12:30..... this is my solution:
code :
$current_time[0] = date('h:i A');
echo "<option>" . $current_time[0] . "</option>";
for($i = 1; $i <=5; $i++) {
    $current_time[$i] = strtotime("+".$step." minutes", strtotime($current_time[0]));
    echo "<option>" . date("h:i A", $current_time[$i]) . "</option>";

Define variable before while loop or as static in loop

Define variable before while loop or as static in loop

By : user3484084
Date : March 29 2020, 07:55 AM
help you fix your problem
(1) Is it good practice/does it make sense to define the variable timeout_x100us as static inside the loop to minimize its scope?
code :
uint16_t i;
for(i=0; !is_interruptflag() && i<TIMEOUT; i++)

  return ;
PHP: What's the difference between static variable inside a for loop and a non-static variable outside the for loop?

PHP: What's the difference between static variable inside a for loop and a non-static variable outside the for loop?

By : Surendran R
Date : March 29 2020, 07:55 AM
Hope this helps The first and second loops appear to do the same thing. This is because, as with any other statically-initialized variable, static means you only initialize it once. It retains its value as you try to access it again. You'll find that the static $foobar will still exist outside the scope of the loop wherever you declare/initialize it; this is due to the nature of PHP and has nothing to do with the static variable.
The difference is made clear only when you attempt to access $foobar before the loop: it won't be declared yet in the second snippet because you only create it within the loop, so you may get an undefined-variable notice.
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