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I learnt that i+=2 is the short-hand of i=i+2. But now am doubting it. For the following code, the above knowledge holds no good:

byte b=0; b=b+2; //Error:Required byte, Found int

The above code is justifiable, as 2 is int type and the expression returns int value.

But, the following code runs fine:

byte b=0; b+=2; //b stores 2 after += operation

This is forcing me to doubt that the += short-hand operator is somewhat more than I know. Please enlighten me.

CᴴᴀZ
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  • See: http://stackoverflow.com/questions/7038012/typecasting-with-shorthand-operators – Kevin Bowersox May 03 '13 at 15:07
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    No use in answering when a good answer exists. Check [this one](http://stackoverflow.com/questions/8710619/java-operator) out. This question is covered there. – Fritz May 03 '13 at 15:08

2 Answers2

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When in doubt, you can always check the Java Language Specification. In this case, the relevant section is 15.26.2, Compound Assignment Operators.

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.

So you were almost correct, except that a cast is added as well. In your case: b+=2; qualifies to b=(byte)(b+2);

CᴴᴀZ
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Antimony
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Coming from a C++ background, I can attempt to claim that the concept is similar. The operator += is defined as a method of the object to its left, passing in the object to its right as an argument. This makes it trivial for the writer of the language to throw in things like typecasting and compatibility fixes. The only thing is, as far as I know, Java doesn't allow the user to overload the operator functions

Dolphiniac
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