In Java if I do the following I get an error
byte b = 50;
b = b * 2; // Error! Cannot assign an int to a byte!
Ok I understood why I got that error .
But now if I do b*=2 I don't get any error. Why?
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Raj
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1I'm assuming because b*=2 is automatically casting the result back as a byte, you could do `b = (byte) (b * 2);` – mike Mar 28 '17 at 12:45
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2Lot of upvotes for a duplicate... – Murat Karagöz Mar 28 '17 at 12:47
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@MuratK. Voting ring? – Erwin Bolwidt Mar 28 '17 at 12:47
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Just in case: although the question was closed, you could still consider accepting one of the answers ... or go forward and delete the question altogether. Your choice. – GhostCat Mar 28 '17 at 13:53
2 Answers
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Because when you make b *= 2; in fact this operation *= will cast your int to byte.
Youcef LAIDANI
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The reason is simply because there are different rules for narrowing conversions for = and *=.
See here for details on widening in general; and then you go here to understand the difference for those *= operations.
You, see
b *= 2
works out as
b = (byte) ( (b) * 2 )
and that narrowing conversion doesn't play a role here.
GhostCat
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