Hi All,
I am not able to get convinsing/authorising
explanation for this right shift problem...it goes
like this
(A) ~0x0 = 0xFFFFFFFF // this is straight forward
(B) (~0x0>>5 ) = ( 0xFFFFFFFF >>5) // right
//shifting by 5 bits both
// 'left and right hand sides'
But seems (B) is INCORRECT statement seeing the print
!
int number ;
number = ( 0xFFFFFFFF >> 5 ) ;
cout << hex << number << endl ;
number = ( ~0x0 >> 5 ) ;
cout << hex << number << endl ;
--------
output >> 7ffffff
ffffffff
What's the rule followed above ?
This behavior is same on SPARC/ Linux/ Windows.
Does that imply that '~' will always return a signed
value and hence this behavior ?
Thanx in Advance :)
-Manish
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