Thanks a lot, it is quite clear now
On Thu, 19 Jun 2003, Barry Bond wrote:
> Emit32/16/8 just copies the 32/16/8-bit argument directly into the
> output buffer with no mapping.
>
> In your example, 0xfb83 is little-endian for bytes 0x83 0xfb, where 0x83
> is the x86 opcode "CMP r/m32, imm8" (whic
Emit32/16/8 just copies the 32/16/8-bit argument directly into the
output buffer with no mapping.
In your example, 0xfb83 is little-endian for bytes 0x83 0xfb, where 0x83
is the x86 opcode "CMP r/m32, imm8" (which Intel documents as opcode "83
/7 ib", and 0xfb is the mod/rm byte, encoding "mod=3",
