Why local variable make as final whether method argument or inside method.
private void add(final int a , final int b) {
final int c = 0;
}
Please anyone clarify.i am searching a lot but i did not find exact answer.
3 Answers
One reason is it prevents you inadvertently changing them. It's good practise since it'll catch some hard-to-spot coding mistakes.
The second reason is that if you're using inner classes, variables referenced from an outer scope need to be declared as final. See this SO answer for more detail.
The problem with final is that it means different things in different contexts. See this discussion for more info.
8 Comments
Niklas B.
IMHO it also decreases readability. Not sure if it's sensible to do that for every parameter and local variable.
ControlAltDel
In regards to the second reason, this is because if they are not final, the inner class cannot know if they have subsequently changed or not
App Kart
is it not compulsory local make as final?
Voo
Personally I don't think I've ever had a single bug that would've been avoided if I had made a local variable final. Imho final is an exceedingly weak construct compared to say
const in C++.Voo
@Arun Where do you get that it's good practice? It's not part of the Oracle style guide as far as I can see and there's clearly no consensus among programmers - actually as you see several people are quite opposed to it. The advantages and disadvantages (visual clutter, doesn't add much) and when it is actually necessary, have all been named - make up your own mind what you prefer.
|
Functional style uses final variables. This is one reason. Another one is closures.
Final variables are the only ones that can be used in closures. So if you want to do something like this:
void myMethod(int val) {
MyClass cls = new MyClass() {
@override
void doAction() {
callMethod(val); // use the val argument in the anonymous class - closure!
}
};
useClass(cls);
}
This won't compile, as the compiler requires val to be final. So changing the method signature to
void myMethod(final int val)
will solve the problem. Local final variable will do just as well:
void myMethod(int val) {
final int val0;
// now use val0 in the anonymous class
Comments
final in these instances simply means that the values cannot be changed. Trying to set another value to any of your final variables will result in a compile-time error
Comments
finalhas absolutely no consequence when adding it to parameters (well apart from the obvious ones that are true in every context).