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doubt shift ??

 
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fnrds

my doubt here is that whenever there is a shift being done ,it is promoted to int ..as here
-64 means 1100 0000
which gets promoted while shifting to int ..
and after which truncated to byte nd evaluated...right??
 
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Hi Srikanth,

What I understand ( and people please correct if I am wrong) is first the byte number is promoted to an integer. In this process, -64 erstwhile represented as 0xc0 would now be represented as 0xffffffc0 (in 32 bit register). Thereafter the number is shifted and the result(-4) is returned as int.
 
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When you perform a shift operation, both operands, the value to be shifted and the amount to shift it by, must be an int or a long.
{
Here in (b >>>4) b=-64 & 4 both needs to be an int or long
}
If they aren't, they are promoted (if they can be � if not, you end up with a compiler error).
{
since byte b=-64.....b is converted to int & shift operation is done then after...
}

The result of the operation is the type of the left hand operand, which will be an int or a long.
{
After shift operation the result is an int....
}

As neither of those types are assignable to a byte, It must cast it as a byte or the compiler will complain.
[ November 07, 2005: Message edited by: Purujit Saha ]
 
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remember viewers: this is a 1.4 topic but NOT a 5.0 topic!
 
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