Operator Oddities in Java

Question

int number = 10;
int result = 0
        for (int i = 10; i <= 50; i += 10) {
            number = i;
            result = --number - number--;
            System.out.println("i = " + i + ", number = " + number +
                    ", result = " + result);

The result is

i = 10, number = 8, result = 0

i = 20, number = 18, result = 0

i = 30, number = 28, result = 0

i = 40, number = 38, result = 0

i = 50, number = 48, result = 0

I want to know how the operators work step by step in result = --number - number-- ; . What are the values of number while executing the above statement? Thank you.

Have you tried using a debugger? They're pretty good at this kind of stuff. — Federico klez Culloca, Feb 23 '21 at 10:51
Does this answer your question? [How do the post increment (i++) and pre increment (++i) operators work in Java?](https://stackoverflow.com/questions/2371118/how-do-the-post-increment-i-and-pre-increment-i-operators-work-in-java) — OH GOD SPIDERS, Feb 23 '21 at 10:51
Java evaluates left-to-right, irrespective of operator precedence. So, it evaluates `--number`, then `number--`, then subtracts the results. — Andy Turner, Feb 23 '21 at 10:52
because its not subtracting 9-10. It is subtracting 9-9. Why do you think `number` is still 10 at that moment when it will already have evaluated `--number` at that point which will have decremented number from its original value? — OH GOD SPIDERS, Feb 23 '21 at 11:03

Arvind Kumar Avinash · Accepted Answer · 2021-02-23T11:08:34.287

2

I think only this much code is enough to make you understand this concept:

public class Main {
    public static void main(String[] args) {
        int number = 10;
        int result = --number - number--;
        System.out.println("number = " + number + ", result = " + result);
    }
}

This is how it is processed:

--number => number = number - 1 => number = 10 - 1 => number = 9 
result = --number - number-- => result = 9 - 9 => result = 0
number-- => number = number - 1 => number = 9 - 1  => number = 8

Thus the output:

number = 8, result = 0

edited Feb 23 '21 at 11:08

answered Feb 23 '21 at 11:02

Arvind Kumar Avinash

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result = number-- - number++ * --number; – Emily Watson Feb 23 '21 at 11:19
Can you explain me internal working of above codes? Thank you, sir. – Emily Watson Feb 23 '21 at 11:20
1

@EmilyWatson - Same as I've explained above, it will be evaluated as `result = 10 - 9 * 9 = - 71` – Arvind Kumar Avinash Feb 23 '21 at 11:25

Andy Turner · Answer 2 · 2021-02-23T12:00:32.853

To understand what --number and number-- mean, it is useful to consider that these are both expressions with side-effects

An expression is a part of the code which has a value.
A side-effect means that something changes as a result of evaluating that expression.

Looking at each of them in turn:

--number's side effect is to decrease the number's value by 1; its value is the number stored in number after decreasing it.
number--'s side effect is to increase the number's value by 1; its value is the number stored in number before decreasing it.

Because Java evaluates expressions left-to-right, in --number - number--, --number is evaluated first; then number-- is evaluated next.

The initial value of number is some value, say, X
The value of --number is X - 1; the value stored in number afterwards is X - 1.
The value of number-- is the same value, X - 1; the value stored in number afterwards is X - 2.
Hence, --number - number-- is (X - 1) - (X - 1), which is zero
The final value of number is X - 2.

This answer helps me to understand.Thank you so much. – Emily Watson Feb 23 '21 at 11:54 — Emily Watson, Feb 23 '21 at 11:54

WJS · Answer 3 · 2021-02-23T13:29:29.803

1

A slightly different explanation. Your code. Let i = 10

number = i;
result = --number - number--;

Is the same as

number = i;
number = number - 1;  // number is now 9
result = number;      // result is now 9
result = number - number;  // result is now 0
number = number - 1;       // number is now 8

so the first line would be

i = 10, number = 8, result = 0

Pre-increments and decrements are applied before using the value, and post increments and decrements are applied after using the value.

edited Feb 23 '21 at 13:29

answered Feb 23 '21 at 13:22

WJS

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This help me alot. Thanks so much. – Emily Watson Feb 23 '21 at 14:47

Operator Oddities in Java

3 Answers3