shift operator

[Jean]

Who can tell me why:
int i=1;
i<<=31;<br /> i>>=31;
the running result is i=-1;
int i=1;
i<<=31;<br /> i>>>=31;
the result is i=1;

[Hima Mangal]

hi Jean..
well.. the result has to do with signed right shift(>>) and unsigned right shift operators(>>>)
before we start on an example, let it be clear that a signed right shift fills the higher-order bits with the sign bit of the number (1 or 0) whereas unsigned right shift fills the higher-order bits with 0s regardless of the sign bit..
1 when represented in binary becomes:
Number Binary Representation(Memory)
i 1 00000000000000000000000000000001
i i<<31 10000000000000000000000000000000 <br /> i i>>31 11111111111111111111111111111110
and this is the 2's complement representation of -1, so i comes out to be -1.
In the second case when we apply the unsigned right shift operator..
i i>>>31 00000000000000000000000000000001
the higher-order bits are filled with 0s instead of 1(as already explained) so, i comes out to be 1.
hope this helps..
------------------
Hima

[Jane Griscti]

Hi Jean,
Please read the JavaRanch Name Policy and re-register using a name that complies with the rules.
Thanks for your cooperation.
------------------
Jane Griscti
Sun Certified Programmer for the Java� 2 Platform

[S Dave]

int i=1;
00000000 00000000 00000000 00000001
i<<=31<br /> 10000000 00000000 00000000 00000000<br /> i>>31 : inserts sign bit 0/1 in shifted bit position
11111111 11111111 11111111 11111111
which is -1 in binary notation
int i=1;
00000000 00000000 00000000 00000001
i<<=31<br /> 10000000 00000000 00000000 00000000<br /> i>>>31 : inserts 0 in shifted bit position
00000000 00000000 00000000 00000001
which is 1 in binary notation

[Jean Kran]

Thank you Hima and Sweekriti,I can understand the shift operator
right now.