byte question

[satellite03]

hi, i am testing this code

byte x =64,y;
y = (byte)(x<<2);
System.out.println(y);

output is 0. why?

but, if i test

byte x =64,y;
y = (byte)(x<<1);
System.out.println(y);

its giving -128 . why ?

i know byte has range max=127 and min = -128 .

but using this two, i find the above two results are contradictory. both the cases results are out of range...but they are printing different results. why?

 

[sedj]

This may help :

http://www.csc.liv.ac.uk/~frans/COMP101/AdditionalStuff/bitOps.html

 

[satellite03]

hi, sedj, thanks for the link. i am aware of the shift operators.

if i want to get the number after shifting, then quick rule is multiply the number by 2^number_of_shifts(for left shift).

so applying this quick rule in my first code

64*2^2 = 256

and in second code,64*2^1 = 128

both of them are out of byte max range 127. is not it?

but while printing, they did not print it, rather they printed different output.
so why they gave output differently , why not the same?



you know, i was expecting -128( which is min value of byte ), bcoz i thought, if it is out of range then it will go back reverse way to -128 and onwards.


i also thought,the output to be the highest value(i.e 127) if it goes out of range....but it is not.

however, those two code printed two different output although both of them are out of range as i have shown.

 

[JavaDude32]

Well in binary 64 is 01000000 and if you shift once then you have 10000000 which acting as a signed number (last bit is the signed bit) then you have -128 by the 2's complement. If you shift twice then you get 00000000 as it does not carry which is obviously 0. You would have to perform an arithmetic operation for overflow to come into effect. Hope that helps.

-JavaDude32