question on bit shifting

[Tualha Khan]

Dear All,
Hey,
Why does the following code print the value of i as 1 and not 0:
int i=1;
i = i >> 32;
System.out.println(i);
What happens to i when it goes past 0 in leftshift,
Also, I have seen in one of the examples that i<<32 will print the result as 1, and i<<33 will result in value 2.
I know that 1<<31 will result in i being set to a negative value which is the maximum value for that type. But what happens after that.
Please clarify!!
Bye,
Tualha Khan

[Velmurugan Periasamy]

Shifting operations never shift more than 31 bits(in case of int) or 63 bits(in case of long). If the RHS operand is more than 31/63, then it is reduced to RHS % 32/64.
So in the following case, no shifting happens since 32 % 32 is zero,so the value still is 1 after the shift.
int i = 1;
i = i >> 32;
In case of i << 33, what really happens is i << 1, so if the value of i was 1 before the shifting, it becomes 2 after the one bit left shifting.
HTH
------------------
Velmurugan Periasamy
Sun Certified Java Programmer
----------------------
Study notes for Sun Java Certification
http://www.geocities.com/velmurugan_p/

[vikas singh]

dear Tualha Khan,
remember %32 whenever you are doing bitwise shifting, take mod 32
of R.H.S operand.
ex :
1 >> 4 // here calculating 4 % 32 = 4 thus 4 bit shifting .
1 >> 31 // here calculating 31 % 32 = 31 thus 31bit shifting .
1 >> 32 // here calculating 32 % 32 = 0 thus 0 bit shifting .
1 >> 33 // here calculating 33 % 32 = 1 thus 1 bit shifting .
during bitwise shifting before any shifting could takes place
the compiler do %32 on RHS operand, which by default is int.
In case it is long than %64 . . .
all the best with bits,
vikas singh .