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", the 7 from the
CMP opcode, and "rm=3", where the mod/rm bits mean "register ebx". The
Emit8(0) that follows copies the value of the imm8 field in the opcode.
The code could have also been written as:
Emit8(0x83);
Emit8(0xfb);
Emit8(0x00);
But the C compiler won't produce as efficient code as it will for
Emit16(0xfb83).
Barry
This posting is provided "AS IS" with no warranties, and confers no
rights.
-----Original Message-----
From: Discussion of the Rotor Shared Source CLI implementation
[mailto:[EMAIL PROTECTED] On Behalf Of Archana
Sent: Thursday, June 19, 2003 3:24 AM
To: [EMAIL PROTECTED]
Subject: [DOTNET-ROTOR] Assembly codes in jitinterfacex86.cpp
Hi,
Is there any documentation regarding the opcodes used in the file
jitinterfacex86.cpp and how the intel opcodes are mapped onto these
codes
for example this is a piece of code from vm/i386/cgencpu.cpp,
// cmp ebx,0
Emit16(0xfb83);
Emit8(0);
if one needs to write
// cmp edx,0
what parameter should be passed to Emit16 etc..
Thanks,
archana
