问题
Consider the following snip of java code
byte b=(byte) 0xf1;
byte c=(byte)(b>>4);
byte d=(byte) (b>>>4);
output:
c=0xff
d=0xff
expected output:
c=0x0f
how?
as b in binary 1111 0001
after unsigned right shift 0000 1111 hence 0x0f but why is it 0xff how?
回答1:
The problem is that all arguments are first promoted to int before the shift operation takes place:
byte b = (byte) 0xf1;
b is signed, so its value is -15.
byte c = (byte) (b >> 4);
b is first sign-extended to the integer -15 = 0xfffffff1, then shifted right to 0xffffffff and truncated to 0xff by the cast to byte.
byte d = (byte) (b >>> 4);
b is first sign-extended to the integer -15 = 0xfffffff1, then shifted right to 0x0fffffff and truncated to 0xff by the cast to byte.
You can do (b & 0xff) >>> 4 to get the desired effect.
回答2:
I'd guess that b is sign extended to int before shifting.
So this might work as expected:
(byte)((0x000000FF & b)>>4)
回答3:
According to Bitwise and Bit Shift Operators:
The unsigned right shift operator ">>>" shifts a zero into the leftmost position, while the leftmost position after ">>" depends on sign extension.
So with b >> 4 you transform 1111 0001 to 1111 1111 (b is negative, so it appends 1) which is 0xff.
回答4:
Java tries to skimp on having explicit support for unsigned basic types by defining the two different shift operators instead.
The question talks about unsigned right shift, but the examples does both (signed and unsigned), and shows the value of the signed shift (>>).
Your calculations would be right for unsigned shift (>>>).
回答5:
The byte operand is promoted to an int before the shift.
See https://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.19
Unary numeric promotion (§5.6.1) is performed on each operand separately. (Binary numeric promotion (§5.6.2) is not performed on the operands.)
And https://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.6.1
Otherwise, if the operand is of compile-time type byte, short, or char, it is promoted to a value of type int by a widening primitive conversion (§5.1.2).
回答6:
byte b=(byte) 0xf1;
if (b<0)
d = (byte) ((byte) ((byte)(b>>1)&(byte)(0x7F)) >>>3);
else
d = (byte)(b>>>4);
First, check the value: If the value is negative. Make one right shift, then & 0x7F, It will be changed to positive. then you can make the rest of right shift (4-1=3) easily.
If the value is positive, make all right shift with >>4 or >>>4. It does'nt make no difference in result nor any problem of right shift.
来源：https://stackoverflow.com/questions/3948220/behaviour-of-unsigned-right-shift-applied-to-byte-variable