Increment in function overwritten by +=

Question

A method called in a ternary operator increments a variable and returns a boolean value. When the function returns false the value is reverted. I expected the variable to be 1 but am getting 0 instead. Why?

public class Main {
    public int a=0;//variable whose value is to be increased in function
    boolean function(){
        a++;
        return false;
    }
    public static void main(String argv[]){
        Main m=new Main();
        m.a+=(m.function()?1:0);
        System.out.println(m.a);//expected output to be 1 but got a 0 !!!!!
    }
}

Lowson

Isn't this supposed to be undefined behavior? – Handspring

Undefined behavior is a C/C++ concept. – Kleiman

dont you think the return type of function must be int – Workout

The expected output should be 0. m.function() always returns false, so the value that gets added on is 0... 0 + 0 = 0. – Brutal

@LukePark but the function increments a – Handspring

@holidayCoder No. – Kleiman

@PatrickRoberts But it doesn't matter that it does. m.a = m.a + 1 or 0 depending on whether function is true. – Brutal

function is returning false which means it is 0 and in operation it is false so m.function()?1:0 also return 0?? – Workout

@holidayCoder false != 0 in Java... Ternary Operator is (condition ? valueIfTrue : valueIfFalse). – Brutal

@LukePark ignoring the information revealed in the answer below, what this looks like it expands to is: a += (a++ == 1) ? 1 : 0; (I know that the function always returns false but this was the only inline statement I could think of as equivalent in this situation. – Handspring

@PatrickRoberts It looks like it does but it doesn't. – Brutal

@LukePark your initial statement is still incorrect though and that is what I am pointing out. The return value of the function and the ternary operator alone have nothing to do with the end value of m.a. It's the fact that the += operator somehow replaces the post-increment state of a. – Handspring

@PatrickRoberts My initial statement isn't wrong. I think you are misunderstanding the order. Replacing values, it's pretty much saying m.a = 0 (the current value) + (potentially 1 or 0). – Brutal

@LukePark my apologies. I misinterpreted your initial claim to ignore the fact that the function incremented a. – Handspring

@PatrickRoberts No worries! – Brutal

What happens if you call instead a += function()?1:0;? The same thing? I don't have a way to compile C# currently. – Kiblah

@PatrickRoberts: Even in C or C++ the equivalent program would not have undefined behaviour. There are sequence points enough to ensure that a well-defined order is imposed. – Howlett

Answer 1

Basically m.a += (m.function() ? 1 : 0) compiles into

 int t = m.a; // t=0 (bytecode GETFIELD)
 int r = m.function() ? 1  : 0; // r = 0 (INVOKEVIRTURAL and, IIRC, do a conditional jump)
 int f = t + r; // f = 0 (IADD)
 m.a = f // whatever m.a was before, now it is 0 (PUTFIELD)

The above behavior is all specified in JLS 15.26.2 (JAVA SE 8 edition)

Contessacontest

Upvoted +1 for you, but -1 for the JLS being foul in this respect. – Purgatory

Answer 2

You have two operations operating on m.a in one call; in main

m.a += (m.function()?1:0);

pushes the value of a on the frame, and then invokes m.function() (which returns false), thus the ternary expands to m.a += 0; (and the value of m.a from the frame is added to 0 and stored in m.a). Thus the value is incremented in m.function() (and then reset in main). Consider it this way,

m.a = m.a + (m.function() ? 1 : 0);

The value of m.a is determined before the evaluation of m.function() (thus it is a post increment operation). For the result you expected, you could do

m.a = (m.function() ? 1 : 0) + m.a;

Cosgrave

Why is it evaluated this way? Can you cite the JLS sources that cover this? – Kleiman

So the behavior is related to stack calls? – Handspring

function is returning false which means it is 0 and in operation it is false so m.function()?1:0 also return 0?? – Workout

I was perhaps unclear, it is really unrelated to the ternary, it is because of the post increment operator being used. – Cosgrave

A prior user answered this with the same information in less detail and people voted him down twice... If he is still here, kudos to you, I always agreed with you. – Brutal

docs.oracle.com/javase/specs/jls/se8/html/… explains how compound assignments are evaluated @JohnKugelman. – Appressed

@holidayCode : yes - m.function() ? 1 : 0 will always return 0 because m.function() will always return false. The entire question is "why is the a++ in function() ignored?". The answer is that the in main() the value of a is pushed onto the stack before function() is called, so the modification to a is not noticed. IMO this is a compiler bug - the compiler should have recognized that a call to a class function could potentially modify a class variable and thus shouldn't have pushed a until after function was completed. – Knockknee