I came across below java line and puzzled about its output. Can you please explain me logic behind this code
System.out.println((int)(char)(byte) -1);
Output:
65535
-
The logic is to confound and confuse. ;) If you ever write code like this its worth commenting why you did it.Peter Lawrey– Peter Lawrey2011-09-21 14:22:34 +00:00Commented Sep 21, 2011 at 14:22
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3 Answers
Well, it's equivalent to:
byte b = -1;
char c = (char) b; // c = '\uFFFF' - overflow from -1
int i = c; // i = 65535
Really the explicit conversion to int in the original is only to make it call System.out.println(int) instead of System.out.println(char).
I believe the byte to char conversion is actually going through an implicit widening conversion first - so it's like this really:
byte b = -1;
int tmp = b; // tmp = -1
char c = (char) tmp; // c = '\uFFFF'
Does that help at all?
5 Comments
JavaGeek
Thanks Jon. I understood that. Just one more doubt. If i put System.out.println((char)(byte) -1); then it's displaying "?". Any reason
Martijn Courteaux
Does this mean
char is unsigned in Java? Which makes it the only unsigned primitve.Martijn Courteaux
@Sukumar: That's because the character
0xFFFF is not recognized by the console.
Jon Skeet
@Martijn: Yes, you're absolutely right about
char being the only unsigned primitive in Java.
jarnbjo
@Jon: Are booleans signed? :)
it's the same as
System.out.println((int) '?');
(byte) -1gives: -1(char) -1gives: ?(int) '?'gives 65535
Comments
In java byte is a signed (twos-complement) 8-bit primitive type. The binary representation of a byte with a value of -1 is 11111111. This then gets cast to a char which is a 16-bit primitive with a value between \u0000 and \uFFFF (0 and 65535) - it would appear that the bits of the byte are left-shifted by 8, with sign-extension. So at this point the binary representation is:
1111111111111111
...or 65535. However, it is then not quite as simple as saying "Oh yes then it is turned into an int so we don't see the character representation and is printed out". In java, all numeric primitives are signed! If we cast the char as a short which is another 16-bit primitive, the program would output -1. However, when we cast it to a 32-bit int. The final binary representation becomes:
00000000000000001111111111111111
...which is 65535 both signed and unsigned!
3 Comments
Voo
In java, all numeric primitives are signed - No. The whole point of this exercise is to show that chars are unsigned in Java. Otherwise the result would be 0xffffffff and NOT 0xffff..
Max Spencer
@Voo Yes,
In java, all numeric primitives are signed, by that I mean all primitives except chars, which do not represent numbers (kind of, maybe that was a bad choice of words..). I don't understand the second part of your repsonse, the result isn't 0xffff?Voo
Ah ok, I'm just used to handle chars like any other primitive without a distinction (too much C?) but I can see how that's valid. The second part just meant that if chars weren't unsigned we'd get 0xffffffff instead of 0xffff.