Why is the result not 1?

Question

int m = 0;

m += m++;

System.out.println(m);

prints to 0 but i thought m will be post incremented and finally be 1. Would someone please explain.

Note: i know how post increment works (atleast i think i do :P). But what i am trying to figure out is that when i say m + = m++ lets assume it means m = m + m++ which will evaluate m++ after sending the value of m, 0 in this case, and evaluate to m = 0 + 0 then increment the value of m because of the post increment. now if post increment has occured, why is m not 1

@kark "i thought m will be post incremented and finally be 1" — eis, Sep 26 '13 at 11:15

score 4 · Answer 1 · edited May 23 '17 at 12:21

4

m++ returns m then increments it.

Change it to ++m and you'll see 1.

For clarity, this question explains it too: What is x after "x = x++"?

edited May 23 '17 at 12:21

Community

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answered Sep 26 '13 at 11:14

christopher

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OP says prints to `0`; – TheKojuEffect Sep 26 '13 at 11:15
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@TheKojuEffect And so it does, and if it is changed as stated here it will print 1. Your point? – user207421 Sep 26 '13 at 11:16
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This doesn't explain what happens with the incremented m in the OP example. – eis Sep 26 '13 at 11:17
@eis Of course it does. "`m++` returns `m` then increments it" explains it completely – user207421 Sep 26 '13 at 11:18
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@EJP And that doesn't explain anything about the question. – TheKojuEffect Sep 26 '13 at 11:18
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@Chris This doesn't answer the question. – Kevin A. Naudé Sep 26 '13 at 11:19
@TheKojuEffect It *answers* the question. – user207421 Sep 26 '13 at 11:19
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i know how post increment works @chris. what i am trying to figure out is that when i say `m + = m++` lets assume it means `m = m + m++` which will evaluate m++ after sending the value of `m`, `0` in this case and evaluate to `m = 0 + 0` then increment the value of `m` because of the post increment. now if post increment has occured, why is `m` not `1` – Thirumalai Parthasarathi Sep 26 '13 at 11:20
@EJP I think he didn't see `+=`, do you? – TheKojuEffect Sep 26 '13 at 11:23
@TheKojuEffect No I don't, actually. I think he's just confused about evaluation order. – user207421 Sep 26 '13 at 11:26
@EJP Besides, OP don't want to print `1` but asking why it is not `0`. – TheKojuEffect Sep 26 '13 at 11:26
@EJP I think you are clear about my point now. – TheKojuEffect Sep 26 '13 at 11:27
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While this answer is not incorrect. It does not explain the underlying reasons behind why m == 0 after it 'should' have been incremented according to someone who doesn't know about evaluation order. – Serdalis Sep 26 '13 at 11:41
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@EJP it does not. If m is incremented, where does the incremented value go? – eis Sep 26 '13 at 11:43

axtavt · Answer 2 · 2013-09-26T11:33:39.380

Consider the following rules:

a += b is equivalent to a = a + b. Therefore m += m++ is equivalent to m = m + m++.
You can see that the first occurence of m at the right sight is evaluated before the increment and produces 0
m++ increments m and returns original value of m before the increment.
So, in your case m++ sets m to 1 and returns 0
Assignment happens after evaluation of the right side, whereas post-increment happens during evaluation of the right side.

Now you can see that your expression is evaluated as follows:

Evaluation of the right side produces 0 and sets m to 1
Assignment sets m to value of the right side (0)

score 2 · Answer 3 · answered Sep 26 '13 at 11:18

2

The sequence of events is:

The RHS is evaluated (0)
The post increment is done (m++).
The evaluated result is assigned (m=0 again).

i.e. this is equivalent to:

tmp = m;
m++;
m = tmp;

If you did m = ++m, the sequence would be:

The pre increment is done (++m).
The RHS is evaluated (1)
The evaluated result is assigned (m=1).

answered Sep 26 '13 at 11:18

SteveP

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+1 since your submission came in before mine. – Kevin A. Naudé Sep 26 '13 at 11:26

score 2 · Answer 4 · answered Sep 26 '13 at 11:22

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ANSWER: An assignment evaluates side effects before performing the assignment.

m += m++;

is equivalent to

temp = m + m++;    ; m is now 1, but temp is 0
m = temp;          ; m is now 0 again

answered Sep 26 '13 at 11:22

Kevin A. Naudé

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score 1 · Accepted Answer · answered Sep 26 '13 at 11:17

int m = 0

At this point m is 0

m += m++;

Expands to m=m+(m++)

m returns 0

m++ returns the value of m which is 0 and then increments it to 1

transforming m+(m++) into 0+0 (m is 1 a this point)

This is then assigned to m resulting in the final value of m.

Tip: avoid post-increment when you're touching the value otherwise

score 1 · Answer 6 · answered Sep 26 '13 at 11:21

1

In your code

m+ =m++; Results -- >> M value + (post incremented m value(m++))

Initially

m value= 0

post incremented(m++) m value = 0 (pre increment(++m) = 1)

answered Sep 26 '13 at 11:21

kark

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score 0 · Answer 7 · answered Sep 26 '13 at 11:15

0

Well m++ return the value of m before to increment it, contrary to ++m that increment then return, like:

 int m++(int m){
     int tmp = m;
     m = m+1;
     return tmp; 
}


 int ++m(int m){
     m = m+1;
     return m; 
}

answered Sep 26 '13 at 11:15

aveuiller

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uh... wat is `m++(int m)` by the way..? – Thirumalai Parthasarathi Sep 26 '13 at 11:18
just a barbarism to represent the function behavior. – aveuiller Sep 26 '13 at 11:21

Peter Sutton · Answer 8 · 2013-09-26T11:26:28.987

f is your code. g is an expanded version showing why they both print 0.

class Test {
    private static void f() {
        int x = 0;
        x += x++;
        System.out.println(x);
    }

    private static void g() {
        int x = 0;
        int preInc = x; // preInc = 0
        x += 1;         // x = 1
        x = preInc;     // x = 0
        System.out.println(x);
    }

    public static void main(String[] args) {
        f();
        g();
    }
}

Why is the result not 1?

8 Answers8