
3 Oct
2008
3 Oct
'08
4:05 p.m.
No. You are invoking undefined behaviour.
Well, by definition you could not be more right, however, in practice my point applies. It is like the x86 sh(l/r) instructions, although for some values larger than the bitfield of the operand the instruction is undefined it behaves in a certain way :D. However, you are right and I am wrong! With best regards Kasra Nassiri