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Modified variable out of function


By : Adam
Date : August 01 2020, 03:00 AM
may help you . This happens because the default arguments b=d is evaluated only once when the function is defined. After that, the value of d inside the function does not change.
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code :


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In class variable cant be modified in function


By : Shyam Patidar
Date : March 29 2020, 07:55 AM
With these it helps This is the code so far: , Try referencing the variable 'hand' through the 'self' pointer:
code :
class Player:
    hand = []
    def take(self, card):
        self.hand.append(card)

variable modified in a function and returned modified when it should not


By : Deborah Mickle
Date : March 29 2020, 07:55 AM
I wish did fix the issue. The problem is that when you assign an existing dictionary to a new variable, actually the new variable points to the same object. You can check basically the address of the object by using id function. In the example, id(calibration) and id(new_calibration) and you will see they have the same address. That's the reason why if the new dictionary is modified also the original dictionary is modified.
In order to have an actual new copy, namely , you can use copy package, and use deepcopy function. The line that you should change is the initialization of new_calibration, namely:
code :
new_calibration = copy.deepcopy(calibration)

VBA variable passed to function is modified


By : Martina Navea
Date : March 29 2020, 07:55 AM
With these it helps Unlike most other languages, the default behaviour for parameters in VBA is for them to be passed ByRef - that means a reference to the original object/variable is passed and thus the original object/variable can be modified by the called function.
The alternative is to use ByVal, in which case a temporary copy of the value is passed to the called function. Because it is temporary, any changes made are then lost when the function ends.
code :
Function Components(ByVal qty As Integer, ByVal stringSize As Integer)

Why isn't the variable in this object modified by its callback function?


By : Hima
Date : March 29 2020, 07:55 AM
should help you out The problem is that setTimeout creates it's own this. Solution may looks like:
code :
start_timer(ptr_callback_function) {
    // savig this that your need
    const self = this;

    setTimeout(function() {
        // use needed context
        self.timer_finished = true;
        ptr_callback_function();
    }, 1000);
}
start_timer(ptr_callback_function) {
    setTimeout(() => {
        this.timer_finished = true;
        ptr_callback_function();
    }, 1000);
}

My function is not printing the modified variable inside it


By : Jason Wu
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , The code is printing the original starting balance of 5000 instead of the modified balance 4900 in the function. , You should return the variable mon.
code :
if building_type=="House":
    # do calculations
    return (mon, house, office)

elif building_type=="Office":
    # do calculations
    return (mon, house, office)
results = Building(money, house, office)
print(results[0]) # for money
print(results[1]) # for house
print(results[2]) # for office
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