John Reiser:
On x86, the byte sequence {0xc7,0310,1,2,3,4} superficially looks like "move
immediate to r/m dword" because of the opcode 0xC7.  Actually, it is an illegal
instruction because 0!=(070 & mod_rm); namely, the 0310 should be 0300.  Gdb
disassembly should report illegal instruction, but instead says "movl
$0x4030201,%eax".

Please see the original post:
https://bugzilla.redhat.com/bugzilla/show_bug.cgi?id=172034

Steps to Reproduce:
1. Compile and run this program under gdb:
-----foo.S
_start: .globl _start
        nop; int3
        .byte 0xc7,0310,1,2,3,4
        nop; nop
-----
$ gcc -o foo -nostartfiles -nostdlib foo.S
$ gdb foo
(gdb) run
2.
3.
  

Actual Results:  Program received signal SIGTRAP, Trace/breakpoint trap.   ## as
expected for 'int3'
0x08048076 in _start ()
(gdb) x/i $pc
0x8048076 <_start+2>:   mov    $0x4030201,%eax   ## gdb says it's OK
(gdb) stepi

Program received signal SIGILL, Illegal instruction.   ## CPU rejects
0x08048076 in _start ()
0x8048076 <_start+2>:   mov    $0x4030201,%eax
(gdb)


Expected Results:  Program received signal SIGTRAP, Trace/breakpoint trap.
0x08048076 in _start ()
(gdb) x/i $pc
0x8048076 <_start+2>:   (bad)  ## modrm of 0310 is illegal for opcode 0xC7