Confused on how java Java Arithmetic operation work in this case

Asked Jul 22 '18 at 14:29

Active Jul 22 '18 at 14:34

Viewed 50 times

I know that during java arithmetic operations data types like byte,short,and char values are automatically widened to int ,I understand why the following example fails to compile :

byte a = 10 ;
byte b = 20 ;          // compile fails
short sum = a + b ;

But what I didn't understand if we modify the preceding example and define variables a and b as final then its compiles successfully :

final byte a = 10 ;
final byte b = 20 ;  // compile successful
short sum = a + b ;

What I didn't understand here is how the key final assure the compiler to do the sum ?

asked Jul 22 '18 at 14:29

e2rabi

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This has been asked and answered several times before, FYI. – T.J. Crowder Jul 22 '18 at 14:31
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@T.J.Crowder Maybe [this one](https://stackoverflow.com/questions/27044674/char-and-byte-with-final-access-modifier-java)? – user1803551 Jul 22 '18 at 14:33
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@user1803551 - Yeah, and [this one](https://stackoverflow.com/questions/1660856/promotion-in-java). – T.J. Crowder Jul 22 '18 at 14:34

Confused on how java Java Arithmetic operation work in this case

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