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# {Tips For Java} Compound assignment operator.

i += j; is not just a shortcut for: i = i + j; But if we try this: int i = 5; long j = 8; Then i = i + j; will not compile but i += j; will compile fine. Because: (JLS) 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. For example, the following code is correct: short x = 3; x += 4.6; and results in x having the value 7 because it is equivalent to: short x = 3; x = (short)(x + 4.6);

3 Answers

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You should be appreciated for posting concepts like this. Keep doing this. I already am aware of this concept but many are not and this will help them.

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Now that's something I didn't know. Thanks for the information

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thank you