Have you tried a newer version of gcc to see what the code looks like?

  Nate

On Mon, Aug 24, 2009 at 10:06 PM, Gabe Black<gbl...@eecs.umich.edu> wrote:
> This appears to be a gcc bug. I will now explain why. If you don't care,
> stop reading. If you do care and you see some place where I'm wrong,
> please, please let me know.
>
>
>
> The interesting part of the function in question disassembles to the
> following:
>
> 0x0000000000d85fc3 <_ZN16SimpleTimingPort10recvTimingEP6Packet+155>:
> mov    0x55ab4e(%rip),%rax        # 0x12e0b18 <curTick>
> 0x0000000000d85fca <_ZN16SimpleTimingPort10recvTimingEP6Packet+162>:
> mov    %rax,%rdx
> 0x0000000000d85fcd <_ZN16SimpleTimingPort10recvTimingEP6Packet+165>:
> add    -0x8(%rbp),%rdx
> 0x0000000000d85fd1 <_ZN16SimpleTimingPort10recvTimingEP6Packet+169>:
> mov    -0x20(%rbp),%rsi
> 0x0000000000d85fd5 <_ZN16SimpleTimingPort10recvTimingEP6Packet+173>:
> mov    -0x18(%rbp),%rdi
> 0x0000000000d85fd9 <_ZN16SimpleTimingPort10recvTimingEP6Packet+177>:
> callq  0xd85d68 <_ZN16SimpleTimingPort15schedSendTimingEP6Packetl>
> 0x0000000000d85fde <_ZN16SimpleTimingPort10recvTimingEP6Packet+182>:
> jmp    0xd85ffb <_ZN16SimpleTimingPort10recvTimingEP6Packet+211>
> 0x0000000000d85fe0 <_ZN16SimpleTimingPort10recvTimingEP6Packet+184>:
> cmpq   $0x0,-0x20(%rbp)
> 0x0000000000d85fe5 <_ZN16SimpleTimingPort10recvTimingEP6Packet+189>:
> je     0xd85ffb <_ZN16SimpleTimingPort10recvTimingEP6Packet+211>
> 0x0000000000d85fe7 <_ZN16SimpleTimingPort10recvTimingEP6Packet+191>:
> mov    -0x20(%rbp),%rax
> 0x0000000000d85feb <_ZN16SimpleTimingPort10recvTimingEP6Packet+195>:
> mov    (%rax),%rax
> 0x0000000000d85fee <_ZN16SimpleTimingPort10recvTimingEP6Packet+198>:
> add    $0x8,%rax
> 0x0000000000d85ff2 <_ZN16SimpleTimingPort10recvTimingEP6Packet+202>:
> mov    (%rax),%rax
> 0x0000000000d85ff5 <_ZN16SimpleTimingPort10recvTimingEP6Packet+205>:
> mov    -0x20(%rbp),%rdi
> 0x0000000000d85ff9 <_ZN16SimpleTimingPort10recvTimingEP6Packet+209>:
> callq  *%rax
> 0x0000000000d85ffb <_ZN16SimpleTimingPort10recvTimingEP6Packet+211>:
> movl   $0x1,-0x24(%rbp)
> 0x0000000000d86002 <_ZN16SimpleTimingPort10recvTimingEP6Packet+218>:
> mov    -0x24(%rbp),%eax
> 0x0000000000d86005 <_ZN16SimpleTimingPort10recvTimingEP6Packet+221>:
> leaveq
> 0x0000000000d86006 <_ZN16SimpleTimingPort10recvTimingEP6Packet+222>:    retq
>
> The part where it has a heart attack is at +209 where it tries to call
> through the value in memory pointed to by %rax. If you look above that a
> few instructions at +191, you'll see where it gets a value off of the
> stack using %rbp, the frame pointer, and puts that into %rax. That value
> is the pointer pkt.
>
> (gdb) p pkt
> $7 = (PacketPtr) 0x1bd6f40
> (gdb) p/x *(uint64_t)($rbp - 0x20)
> $10 = 0x1bd6f40
>
> Because pkts are reference counting pointers, %rax actually points to a
> structure that contains the pointer to the real packet. The instruction
> at +202 removes that level of indirection. Next, the line at +198 adds 8
> to %rax, making it point to the vtable corresponding to the Printable
> base class. You can see that here after all the static members.
>
> (gdb) p *pkt
> $11 = {<FastAlloc> = {_vptr.FastAlloc = 0x1bd7060, static Max_Alloc_Size
> = 512, static Log2_Alloc_Quantum = 3, static Alloc_Quantum = 8, static
> Num_Buckets = 65, static Num_Structs_Per_New = <optimized out>, static
> freeLists = {0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x2912c50, 0x0,
> 0x1bcf358, 0x2b7e8f0, 0x1bd61a0,
>      0x1bd6f40, 0x0 <repeats 52 times>}}, <Printable> =
> {_vptr.Printable = 0xdd7d70}, static PUBLIC_FLAGS = <optimized out>,
> static PRIVATE_FLAGS = <optimized out>, static COPY_FLAGS = 15, static
> SHARED = 1, static EXPRESS_SNOOP = 2, static SUPPLY_EXCLUSIVE = 4,
> static MEM_INHIBIT = 8, static VALID_ADDR = 256,
>  static VALID_SIZE = 512, static VALID_SRC = 1024, static VALID_DST =
> 2048, static STATIC_DATA = 4096, static DYNAMIC_DATA = 8192, static
> ARRAY_DATA = 16384, flags = {_flags = 3840}, cmd = {static commandInfo =
> 0x12e6080, cmd = MemCmd::MessageResp}, req = 0x2b7e8f0, data = 0x0, addr
> = 11529215046068469760,
>  size = 4, src = 0, dest = 8, origCmd = {static commandInfo =
> 0x12e6080, cmd = MemCmd::MessageReq}, time = 231966339456, finishTime =
> 231966444000, firstWordTime = 231966445000, static Broadcast = -1,
> senderState = 0x0}
>
> To make sure it's pointed at the right thing,
>
> (gdb) p/x *(uint64_t *)((uint8_t *)pkt + 8)
> $13 = 0xdd7d70
>
> Next, we can see that %rax is again dereferenced at +202. This is
> extracting the pointer to the virtual destructor of Printable from its
> vtable.
>
> (gdb) x/gx *(uint64_t *)((uint8_t *)pkt + 8)
> 0xdd7d70 <_ZTV9Printable+16>:   0x00000000004e1f74
>
> (gdb) disassemble 0x00000000004e1f74
> Dump of assembler code for function ~Printable:
> 0x00000000004e1f74 <~Printable+0>:      push   %rbp
> 0x00000000004e1f75 <~Printable+1>:      mov    %rsp,%rbp
> 0x00000000004e1f78 <~Printable+4>:      sub    $0x10,%rsp
> 0x00000000004e1f7c <~Printable+8>:      mov    %rdi,-0x8(%rbp)
> 0x00000000004e1f80 <~Printable+12>:     mov    $0xdd7d70,%edx
> 0x00000000004e1f85 <~Printable+17>:     mov    -0x8(%rbp),%rax
> 0x00000000004e1f89 <~Printable+21>:     mov    %rdx,(%rax)
> 0x00000000004e1f8c <~Printable+24>:     mov    $0x0,%eax
> 0x00000000004e1f91 <~Printable+29>:     test   %al,%al
> 0x00000000004e1f93 <~Printable+31>:     je     0x4e1f9e <~Printable+42>
> 0x00000000004e1f95 <~Printable+33>:     mov    -0x8(%rbp),%rdi
> 0x00000000004e1f99 <~Printable+37>:     callq  0x409340 <_zd...@plt>
> 0x00000000004e1f9e <~Printable+42>:     leaveq
> 0x00000000004e1f9f <~Printable+43>:     retq
> End of assembler dump.
>
> Now %rax holds the value 0xdd7d70, the pointer to the Printable vtable
> plus offset 0 which holds the pointer to the desctructor.
>
> (gdb) info registers
> rax            0xdd7d70 14515568
> rbx            0x1731f10        24321808
> rcx            0x2d43c20        47463456
> rdx            0xc      12
> rsi            0x60     96
> rdi            0x1bd6f40        29192000
> rbp            0x7fff2cbc0fd0   0x7fff2cbc0fd0
> rsp            0x7fff2cbc0fa0   0x7fff2cbc0fa0
> r8             0x0      0
> r9             0x0      0
> r10            0x1bc7f30        29130544
> r11            0x7fff2cbc0cf0   140733943909616
> r12            0x7f5824b4ecb0   140016549686448
> r13            0x1bd3f80        29179776
> r14            0x1731f10        24321808
> r15            0x7f58243844a0   140016541516960
> rip            0xd85ffb 0xd85ffb <SimpleTimingPort::recvTiming(Packet*)+211>
> eflags         0x10202  [ IF RF ]
> cs             0x33     51
> ss             0x2b     43
> ds             0x0      0
> es             0x0      0
> fs             0x0      0
> gs             0x0      0
> fctrl          0x37f    895
> fstat          0x0      0
> ftag           0xffff   65535
> fiseg          0x0      0
> fioff          0x0      0
> foseg          0x0      0
> fooff          0x0      0
> fop            0x0      0
> mxcsr          0x1fa0   [ PE IM DM ZM OM UM PM ]
>
> The pkt pointer is then put into %rdi, I believe to act as the "this"
> pointer, and the value pointed to by %rax is called.
>
> Almost all of this is correct so far, but this is the point where things
> break.
>
> If we look at the encoding for the call instruction, we get the following:
>
> (gdb) x/3b (_ZN16SimpleTimingPort10recvTimingEP6Packet+209)
> 0xd85ff9 <_ZN16SimpleTimingPort10recvTimingEP6Packet+209>:      0xff
> 0xd0    0xc7
>
> Looking in table A-2 of AMD manual 3, we see that 0xff is the one byte
> opcode that encodes a group 5 instruction. We now need to look at the
> following modrm byte, 0xd0. That byte breaks down as mod=3, reg=2, and
> r/m=0. Looking at table A-6, we see that a reg field of 2 encodes a CALL
> instruction with an Ev argument. Looking in the operand syntax notation
> key at the top of A.1, E is for a general purpose register or memory
> operand specified by the ModRM byte. Looking at table A-15, we can see
> that with a mod field of 3, the operand is always a register value, not
> a the location pointed to by the register value.
>
> What that ultimately seems to mean is that gcc is using a mod value of 3
> instead of, for instance, 0, and is inadvertently trying to execute the
> vtable of Printable instead of the function it points to. That piece of
> memory is apparently marked no execute, so the program fortunately dies
> instead of going bananas. gdb is also apparently in on it too, and
> disassembles the call instruction to look like it's dereferencing %rax
> when it isn't.
>
> I would very much appreciate it if someone would explain to me why I'm
> wrong since it would be much easier to fix M5 than gcc. Failing that,
> hopefully somebody can get a hold of someone that can actually do
> something about this.
>
> Gabe
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>
>
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