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Variable used, golobal or local?


Variable used, golobal or local?

By : user3099896
Date : January 11 2021, 05:14 PM
Does that help When you declare a variable inside a function it shadows the variable from the outer scope that has the same name. Once you declared x inside the function it will become the x variable for all the commands in the function, even before the outer scope assignment where x = 10.
If you want to write to the x variable in the outer scope you should declare it as global x, that is
code :
x = 10
y = 10
def some():
    global x
    x = 100-x+x
    print(x)
some()

100


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Why Local Inner class requires local variable Final? This Variable is in same method in which Local Inner class define

Why Local Inner class requires local variable Final? This Variable is in same method in which Local Inner class define


By : Maria Jose Atienza G
Date : March 29 2020, 07:55 AM
seems to work fine Because an inner class gets a copy of the environment which is available to it when it is instantiated.
If you don't declare z as final you could have a situation like the following:
code :
  String z = "variable";

  class MyInner {
        public void seeOuter(){
            System.out.println("Outer x is "+x);
            System.out.println("Local variable z is "+z);
        }
    }
  MyInner my=new MyInner();

  z = "another value";

  my.seeOuter();
performance or optimization in Android : final local variable vs local variable

performance or optimization in Android : final local variable vs local variable


By : user3385766
Date : March 29 2020, 07:55 AM
seems to work fine I don't think there are any performance difference between declare final and local.We declare a variable final when we don't want someone override this variable that is you cannot change the value of final variable(It will be constant).This optimization applies only to primitive types and String constants, not arbitrary reference types. Still, it's good practice to declare constants static final whenever possible.
Lambda local variable default value fails to use the closured local variable

Lambda local variable default value fails to use the closured local variable


By : SeaFlyer
Date : March 29 2020, 07:55 AM
seems to work fine Consider the following piece of code:
code :
λ = ->(x = x) {…}
" Use of unassigned local variable" error for three local variables but not for the fourth local variable. Why

" Use of unassigned local variable" error for three local variables but not for the fourth local variable. Why


By : Jackson Nguyen
Date : March 29 2020, 07:55 AM
it fixes the issue For the following code in C#, why are the errors obtained: - , Your problem is here:
Why is there no compile time error for not initializing local variable if a local variable is declared like this int a=1

Why is there no compile time error for not initializing local variable if a local variable is declared like this int a=1


By : h3ll bouy
Date : January 02 2021, 06:48 AM
may help you . As far as the Java compiler is concerned, uninitialized local variables are fine. As long as you don’t try to use their value, that is. Only your IDE should warn you that m is not used. My Eclipse says “The value of the local variable m is not used”. It’s just a warning, nothing that stops your program from running.
It’s when you try to use the (undefined) value, as in your second snippet, that the error is reported.
code :
        int a=10,b=3,m;
        System.out.println("Hello World "+a+" " + b);
        m = 14;
        System.out.println(m);
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