Why don't Java's +=, -=, *=, /= compound assignment operators require casting?

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暖寄归人
暖寄归人 2020-11-21 05:06

Until today, I thought that for example:

i += j;

Was just a shortcut for:

i = i + j;

But if we try this:<

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  •  栀梦
    栀梦 (楼主)
    2020-11-21 05:21

    Java Language Specification defines E1 op= E2 to be equivalent to E1 = (T) ((E1) op (E2)) where T is a type of E1 and E1 is evaluated once.

    That's a technical answer, but you may be wondering why that's a case. Well, let's consider the following program.

    public class PlusEquals {
        public static void main(String[] args) {
            byte a = 1;
            byte b = 2;
            a = a + b;
            System.out.println(a);
        }
    }
    

    What does this program print?

    Did you guess 3? Too bad, this program won't compile. Why? Well, it so happens that addition of bytes in Java is defined to return an int. This, I believe was because the Java Virtual Machine doesn't define byte operations to save on bytecodes (there is a limited number of those, after all), using integer operations instead is an implementation detail exposed in a language.

    But if a = a + b doesn't work, that would mean a += b would never work for bytes if it E1 += E2 was defined to be E1 = E1 + E2. As the previous example shows, that would be indeed the case. As a hack to make += operator work for bytes and shorts, there is an implicit cast involved. It's not that great of a hack, but back during the Java 1.0 work, the focus was on getting the language released to begin with. Now, because of backwards compatibility, this hack introduced in Java 1.0 couldn't be removed.

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