Anomaly in summing two byte variables in Java

Question

Please check the following peices:

byte a=1, b=2;  b+=a;

is perfectly legal. While,

byte a=1, b=2;  b=b+a;

is not allowed. Although both of these are considered equivalent. Is there any difference in the mechanism of the two ways of assignment?

@DaveNewton oh no, I looked it up in the spec, but had better looked it up on SO ;) — Vincent van der Weele, Jun 22 '13 at 12:53

score 1 · Accepted Answer · answered Jun 22 '13 at 12:52

From the language specification:

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T)((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.

This means that b+=a is equivalent to (byte)(a+b), instead of simply a+b. The latter still needs a cast to byte.

Anomaly in summing two byte variables in Java

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