Java: byte cast issue

Question

I have the following code:

byte b=10;
System.out.println("Test b:"+b);

This codes is compiled and works without problems. However, I cant understand why. 10 is literal here and it is integer literal (default for integer). So we have here casting as on the left side we have byte and on the right side we have integer. And this is narrowing as byte<integer. As I understand narrowing must always be explicit casting otherwise the code won't compile. Could anyone explain why this code works?

`long x = 1;` this compiles too. but `int x = 100000000000000;` wont — Uma Kanth, Aug 19 '15 at 06:54
[Here](http://stackoverflow.com/a/12419618/1343161) is a nice explanation. — Keppil, Aug 19 '15 at 06:55

score 4 · Accepted Answer · answered Aug 19 '15 at 06:54

4

It works, because 10 is a compile-time constant. The compiler can infer that it fits the byte range, so it allows the assignment.

You can a play a bit with assigning values to a byte variable. If you for example do

 byte b = 128;

then you'll get a compilation error, because 128 doesn't fit the byte range. In this case you can do a cast:

 byte b = (byte) 128;

but then you'll end up with an overflow and b will be evaluated to -128.

answered Aug 19 '15 at 06:54

Konstantin Yovkov

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Then could you explain why float f=23.22 not work? 23.22 also fits float range... – Pavel_K Aug 19 '15 at 07:11
The floating point values are something else - checkout the [JLS](https://docs.oracle.com/javase/specs/jls/se7/html/jls-4.html#jls-4.2.3) – Konstantin Yovkov Aug 19 '15 at 07:14

score 0 · Answer 2 · edited Aug 19 '15 at 07:03

0

byte's max value is 127.

10 fits in the byte so it is a valid assignment.

try

byte b = 140; //it wont

edited Aug 19 '15 at 07:03

Konstantin Yovkov

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answered Aug 19 '15 at 06:56

Uma Kanth

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Java: byte cast issue

2 Answers2