Java Initialize Variable Incremented

Question

Is it risky to initialize a global variable with an increment of another global variable?

Example:

int a=0; 
int b=a++; 
int c=a++; 
int d=a++;

This should output:

0,1,2,3

Could it happen that compiler could read a global value before the other?

There shouldn't be a problem with doing this but a will end up being 3 not 0, since you are still incrementing it. Also b will start at 0 since you are effectively assigning then incrementing. But I don't see why you would need to, if you're giving a variable initial data you obviously know the initial data you want! ^^ — Callum.
– Callum., Commented Apr 29, 2015 at 15:20
Perhaps the developer will later want to change the original value. — Cristik
– Cristik, Commented Apr 29, 2015 at 15:20
But in that case it would still be just as easy to type out the new values (providing all he wanted to do was still increment) — Callum.
– Callum., Commented Apr 29, 2015 at 15:21

Luiggi Mendoza · Accepted Answer · 2015-04-29 15:17:31Z

4

It will behave as expected. If you try to use a field before it's defined, the compiler will throw an error:

public class Foo {
    int a = b++; //compiler error here
    int b = 0;
}

This is covered in JLS 8.3

For your case, the output of the variables if they're not modified, would be:

a = 3
b = 0
c = 1
d = 2

answered Apr 29, 2015 at 15:17

Luiggi Mendoza

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Comments

talex · Accepted Answer · 2015-05-21 09:42:51Z

3

Result will be a=3, b=0, c=1, d=2.

If all this variable declared in one class they will be initialized in order of occurrence in code.

PS: b = 0 because a++ get value and then increment variable.

answered Apr 29, 2015 at 15:21

talex

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