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[Qemu-devel] understanding how arpl is translated
From: |
Jun Koi |
Subject: |
[Qemu-devel] understanding how arpl is translated |
Date: |
Tue, 13 May 2008 12:32:00 +0900 |
Hi,
I am trying to understand how "arpl" insn (i386) is translated. In
translate.c we have:
.....
modrm = ldub_code(s->pc++);
reg = (modrm >> 3) & 7;
mod = (modrm >> 6) & 3;
rm = modrm & 7;
if (mod != 3) {
gen_lea_modrm(s, modrm, ®_addr, &offset_addr);
gen_op_ld_T0_A0(ot + s->mem_index); // (1) ****
} else {
gen_op_mov_TN_reg(ot, 0, rm); // (2) ****
}
if (s->cc_op != CC_OP_DYNAMIC)
gen_op_set_cc_op(s->cc_op);
gen_op_arpl();
s->cc_op = CC_OP_EFLAGS;
...
I can see that we decrypt 2 operands of arpl and then call
gen_op_arpl(). This function finally leads to execute op_arpl(), which
is defined as:
void OPPROTO op_arpl(void)
{
if ((T0 & 3) < (T1 & 3)) {
/* XXX: emulate bug or 0xff3f0000 oring as in bochs ? */
T0 = (T0 & ~3) | (T1 & 3);
T1 = CC_Z;
} else {
T1 = 0;
}
FORCE_RET();
}
Obviously op_arpl() relies on T0 and T1 have the value of the 1st and
2nd operands of the above "arpl" insn. However, I can only see that we
copy the 1st operand into T0 at (1) or (2) in the first snippet, but I
never see when we copy 2nd operand into T1. This confuses me, or I
missed something here?
Many thanks,
Jun
- [Qemu-devel] understanding how arpl is translated,
Jun Koi <=