k&b Operators and Assignment Doubt

[anand phulwani]

Hi Ranchers,

On page 148,Ch-3 ,Operators and assignment it is given that

byte b=27;
but only....,the compiler puts in the cast,the preceding code is identical to the following
byte b=(byte)27;

Now on page 149,last line

byte a=128;//byte can only hold up to 127

And continuing on Page 150,

We can fix it with the cast
byte a=(byte)128;

Ok..,i know that
byte b=27;
byte b=(byte)128;
will compile

but this will not
byte b=128;

but according to the first quote mentioned from k&b,as the book says,doing
byte b=128;
should also have an implicit cast like byte b=(byte)128; when
byte b=27;
is having an implicit cast.
and thus byte b=128; should also compile and run.

i know cast will not be needed in the range of the primitive but here
in the book they have not mentioned it???

Thanks,
Anand

[Kunal Mehrotra]

Hi Anand,

The compiler doesnt narrow down a value unless you mention it otherwise which requires *explicit* type casting to truncate bits. If without tinkering with the *value*, the compiler can assign an *int* literal to a *byte* then it would go a head and do it without cribbing but if the left hand literal doesnt fit in the Right hand variable then it would throw a compilation error in doing the same.

Hope this help.
Kunal

[Santana Iyer]

three condition for implicit narrowing conversion during assignment

source is of byte, short, char or int
destination is of type byte,short,char
value of source is in range with destination at compile time

if all above conditions met than imlicit narrowing conversion allowed at assignment time.