Re: [tip:x86/asm] x86/entry/64: Add two more instruction suffixes

[Denys Vlasenko <dvlasenk@redhat.com>]

On 07/03/2018 10:46 AM, David Laight wrote:
> From: Jan Beulich
>> Sent: 03 July 2018 09:36
> ...
>> As said there, omitting suffixes from instructions in AT&T mode is bad
>> practice when operand size cannot be determined by the assembler from
>> register operands, and is likely going to be warned about by upstream
>> gas in the future (mine does already).
> ...
>> -	bt	$9, EFLAGS(%rsp)		/* interrupts off? */
>> +	btl	$9, EFLAGS(%rsp)		/* interrupts off? */
> 
> Hmmm....
> Does the operand size make any difference at all for the bit instructions?
> I'm pretty sure that the cpus (386 onwards) have always done aligned 32bit
> transfers (the docs never actually said aligned).
> I can't remember whether 64bit mode allows immediates above 31.

Immediates up to 63 are allowed in 64 bit mode (IOW: for REX-prefixed form)
(run-tested).

Keep in mind that this instruction is "special" with register bit offset:

Register/memory form (BT REG,[MEM]) does not limit or mask the value of bit offset
in REG, the instruction uses bit REG%8 in byte at address [MEM+REG/8].

This works correctly even for negative values: REG = -1 will access
the most significant bit in the byte immediately before MEM.

Thus, for accesses of standard RAM locations (not memory-mapped IO and such),
the "operand size" concept for this instruction (and BTC, BTR, BTS)
does not make much sense: it accesses one bit. The width of actual memory
access is irrelevant.

I'd say assembler should just use the "natural" width for current mode
(16 or 32-bit), and warn when code tries to use immediate operand which
will be truncated and thus needs a wider operand size.

Intel documentation says that immediate operand in BT IMM,[MEM]
is truncated to operand size. My experiment seems to confirm it:

254:1  <- BT 254,[MEM] actually accesses bit #30, not #254
255:0
254:0
255:0

#include <string.h>
#include <stdio.h>
int main(int argc, char **argv, char **envp)
{
         char buf[256];
         int result;

         memset(buf, 0x55, sizeof(buf)); /* bit pattern: 01010101 */

         buf[255/8] = 0;
         asm("\n"
         "       bt      %1, %2\n"
         "       sbb     %0, %0\n"
         : "=r" (result)
         : "i" (254), "m" (buf)
         );
         printf("254:%x\n", !!result);
         asm("\n"
         "       bt      %1, %2\n"
         "       sbb     %0, %0\n"
         : "=r" (result)
         : "i" (255), "m" (buf)
         );
         printf("255:%x\n", !!result);
         buf[255/8] = 0x55;

         buf[31/8] = 0;
         asm("\n"
         "       bt      %1, %2\n"
         "       sbb     %0, %0\n"
         : "=r" (result)
         : "i" (254), "m" (buf)
         );
         printf("254:%x\n", !!result);
         asm("\n"
         "       bt      %1, %2\n"
         "       sbb     %0, %0\n"
         : "=r" (result)
         : "i" (255), "m" (buf)
         );
         printf("255:%x\n", !!result);

         return 0;
}

When I use "r" instead of "i" to generate REG,[MEM] form,
the instruction does access bit #254:

254:0
255:0
254:1
255:0