From: Andrew Cooper > Sent: 09 June 2023 00:58 > ... > The important point is the l suffix on btsl, which forces it to be long > (32bit) irrespective of the default operand size. Does that matter at all? The 'bit' opcodes (I'm sure 'bts' is 'bit test and set') take a bit offset from the base address. This accesses the same bit regardless of the operand size. The one real issue is that a byte operand will only lock the one byte. This might be problematic if non-bit locked accesses are also used. Although it would need to be rather obscure use. (This may be one of them...) The only other problem is that btsl always does a locked 32bit access. If the base address is misaligned this is a misaligned locked access - problematic if it crosses a cache line boundary. David - Registered Address Lakeside, Bramley Road, Mount Farm, Milton Keynes, MK1 1PT, UK Registration No: 1397386 (Wales)
