Re: Q about increment and assignment


"John W. Kennedy" <jwkenne@xxxxxxxxxxxxx> wrote in message
news:c0ijh.1967$Bf.27@xxxxxxxxxxxxxxx
Big D wrote:
I'm confused by the output of the following code:

public class PP {
public static void main(String[] args) {
int i = 1;
i = i++;
System.out.println(i);
}
}

It outputs 1.

I understand the assignment operator happening before the ++,
but I don't understand why the ++ doesn't increment. I thought
the statement should basically expand to:

i = i;
i = i+1;

but the ++ gets lots somewhere...

No, it gets expanded into:

t = i;

John, is t a copy of the bits in i made by the postfix ++ method?

i = i + 1;
i = t;

Is it fair then to say that t gets restored because of the way the
postfix operator applies only after the expression is evaluated?

What is the difference between the original problem and this one?

int i=1;
System.out.println("i=" + i++); //output=1
System.out.println("i=" + i);//output=2

I find this one easier to understand because eventually the new
bits do output. They don't get overwritten. In the original problem
there is overwriting going on that cause the 2 value to be lost
entirely. Is this due to the self-referencing aspect of the
original problem i=i++;?

One last question: Where is that temp copy of the bits visible? I
can't see it in a debugger, so maybe I'm not using it (Eclipse) to
great advantage. I see the original i bits on the stack and that's
about it. This should not involve heap storage, so that leaves me
wondering where/how I'd ever see that temp set of bits.

thanks ...






.

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