* Jeremy Fitzhardinge ([email protected]) wrote:
> H. Peter Anvin wrote:
> > Allowing different registers should be doable, but if so, one would have
> > to put 0: at the *end* of the instruction and use (0f)-4 instead, since
> > the non-%eax forms are one byte longer.
> >
>
> OK, that's already a problem since its using "=r" as the constraint.
>
> > This also seems "safer", since an imm32 is always the last thing in the
> > instruction.
>
> Good idea. If gas/gcc generates entirely the wrong addressing mode,
> then we've got bigger problems.
>
Ok, let's have a good look at what we want:
1 - get a pointer to the beginning of the immediate value within the
instruction.
2 - make sure that the immediate value, within the instruction, is
written to atomically wrt all CPUs, even on older architectures
where non aligned writes are not atomic.
Effectively, placing a label at the end of the instruction, and then
offsetting backward from there, will give us (1).
Then, for the alignment, we have to give a good look at the instruction
set reference, mov instruction, to see what the variants of mov
immediate value to register are on i386.
First, let's look at the possible prefixes:
- Lock and repeat prefixes : no
- Segment override prefixes: no memory reference there, so doesn't
apply.
- Branch hints : no, it's a mov instruction
- Operand-size override prefix: *yes*, can be used for 2 bytes mov
- Address-size override prefix: no address in there, only immediate
value and register.
(looking at the Compat/Leg Mode)
3 cases:
* 1 byte
B0 + rb MOV r8, imm8 (1 byte opcode)
REX + B0 + rb MOV r8, imm8 (only on 64 bits archs, never generated)
C6 /0 MOV r/m8, imm8 (2 bytes opcode)
(this one doesn't require alignment at all, since we do a 1 byte write)
* 2 bytes
B8 + rw MOV r16, imm16 (1 byte opcode)
66 B8 + rd MOV r16, imm16 (2 bytes opcode) (with 66H prefix)
C7 /0 MOV r/m16, imm16 (2 bytes opcode)
(Alignment on 4 bytes boundaries would be required because of the
possible 1 byte opcode ? Or is the 66H prefix mandatory there ? If it
is, then we can safely align on 2 bytes boundaries.)
* 4 bytes
B8 + rd MOV r32, imm32 (1 byte opcode)
C7 /0 MOV r/m32, imm32 (2 bytes opcode)
(the 2 bytes opcode can be a problem)
I have missed anything ?
Mathieu
--
Mathieu Desnoyers
Computer Engineering Ph.D. Student, Ecole Polytechnique de Montreal
OpenPGP key fingerprint: 8CD5 52C3 8E3C 4140 715F BA06 3F25 A8FE 3BAE 9A68
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