Java assignment operator behavior vs C++

Question

This occured while I was tackling a 'Cracking the Coding interview' question:

Write a function to swap a number in place (that is, without temporary variables)

I decided to write up my solution in Java (because I plan on using Java in my internship interviews.)

I came up with a solution that I was almost confident was the right answer (because I did it in one line):

    public static void main(String args[]) {
        int a = 5;
        int b = 7;
        a = b - a + (b = a);
        System.out.println("a: " + a + " b: " + b);
    }

Surely enough, this code executes the desired result. a == 7 and b == 5.

Now here's the funny part.

This code won't run in C++ nor is this solution in the back of the book.

So my question is: Why exactly does my solution work? I am assuming Java does things differently than other languages?

I think (b = a) doesn't return anything in C++. Or maybe I'm wrong, I don't know C++. — Sweeper
– Sweeper, Commented Sep 17, 2015 at 5:30
@Sweeper b = a is evaluated to an lvalue of b, with the side effect of assignment. — user2486888
– user2486888, Commented Sep 17, 2015 at 5:32
@Sweeper As an expression, it yields the value assigned .. however, C/C++ have the 'fun' issue of sequence points and thus Undefined Behavior. In Java it's defined behavior, albeit 'too clever for me'. — user2864740
– user2864740, Commented Sep 17, 2015 at 5:33
Search for and read about sequence points (I also recommend this evaluation-order reference). — Some programmer dude
– Some programmer dude, Commented Sep 17, 2015 at 5:34

Erwin Bolwidt · Accepted Answer · 2015-09-17 06:16:52Z

5

Looks at the Java Language Specification, section 15.7 (Evaluation Order):

The Java programming language guarantees that the operands of operators appear to be evaluated in a specific evaluation order, namely, from left to right.

So in Java the evaluation order is well-defined and does what you expect.

The C++ specification doesn't provide such a guarantee; in fact it's Undefined Behavior so the program could literally do anything.

Quoting from the cppreference, noting that no sequencing rule exists for sequencing operands to arithmetic operators:

If a side effect on a scalar object is unsequenced relative to a value computation using the value of the same scalar object, the behavior is undefined.

edited Sep 17, 2015 at 6:16

answered Sep 17, 2015 at 5:36

Erwin Bolwidt

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1 Comment

Amadan Over a year ago

In contrast, "If a side effect on a scalar object is unsequenced relative to a value computation using the value of the same scalar object, the behavior is undefined." (cppreference), noting that no sequencing rule exists for sequencing operands to arithmetic operators.

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