c++cx86profilingintel-pin

intel Pin: analysis routine detects ah register instead of rsp (REG_STACK_PTR)

I asked this question few days ago.

I wanted to get the stack allocation size (after the function creation). The answer suggests to do:

if((INS_Opcode(ins) == XED_ICLASS_ADD || INS_Opcode(ins) == XED_ICLASS_SUB) && 
   REG(INS_OperandReg(ins, 0)) == REG_STACK_PTR && INS_OperandIsImmediate(ins, 1)

Which in theory is correct and does make sense. But, it doesn't work in practice (correct me if I'm wrong here). It works perfectly fine if I remove REG(INS_OperandReg(ins, 0)) == REG_STACK_PTR check. Why? Because pin doesn't detect the REG_STACK_PTR register when REG(INS_OperandReg(ins, 0)) is used to detect it. rather, it detects ah (which I believe is RAX), when I do check against add rsp, 0xffffffffffffff80 instruction (so, every time it gives: register: ah), as can be seen in my output below:

in
register: rbp
40051e  push rbp
register: *invalid*
value: -128
40051f  mov rbp, rsp
register: ah
400522  add rsp, 0xffffffffffffff80
register: *invalid*
400526  mov dword ptr [rbp-0x28], 0x7
register: *invalid*
40052d  mov dword ptr [rbp-0x64], 0x9
register: eax
400534  mov eax, 0x0
register: *invalid*
400539  call 0x4004e6
register: rbp
4004e6  push rbp
register: *invalid*
value: 64
4004e7  mov rbp, rsp
register: ah
4004ea  sub rsp, 0x40
register: *invalid*
4004ee  mov dword ptr [rbp-0xc], 0x4
register: rax
4004f5  lea rax, ptr [rbp-0xc]
register: *invalid*
4004f9  mov qword ptr [rbp-0x8], rax
register: rax
4004fd  mov rax, qword ptr [rbp-0x8]
register: eax
400501  mov eax, dword ptr [rax]
register: *invalid*
400503  mov esi, eax
register: edi
400505  mov edi, 0x4005d0
register: eax
40050a  mov eax, 0x0
register: rdi
40050f  call 0x4003f0
register: rdi
4003f0  jmp qword ptr [rip+0x200c22]
register: *invalid*
4003f6  push 0x0
register: *invalid*
4003fb  jmp 0x4003e0
register: *invalid*
4003e0  push qword ptr [rip+0x200c22]
register: rdi
4003e6  jmp qword ptr [rip+0x200c24]
4
register: *invalid*
400514  mov dword ptr [rbp-0x3c], 0x3
40051b  nop
register: *invalid*
40051c  leave 
register: *invalid*
40051d  ret 
register: eax
40053e  mov eax, 0x0
register: *invalid*
400543  leave 
out

Well, interestingly it does this for every occurrences of rsp (i.e. it detects ah instead of rsp). Also, it always prints the instruction 400522 add rsp, 0xffffffffffffff80, including rsp (So, why it doesn't print ah here?)

If ah represents rsp in some way, then I can always detect ah using: REG(INS_OperandReg(ins, 0)) == REG_AH. But, I want to understand what is going on here.

My code:

#include <iostream>
#include <fstream>
#include "pin.H"
#include <unordered_map>

// key to open the main Routine
static uint32_t key = 0;

// Ins object mapping
class Insr
{
private:
  // Disassembled instruction
    string insDis;
  INS ins;

public:
    Insr(string insDis, INS ins) { this->insDis = insDis; this->ins = ins;}
    string get_insDis() { return insDis;}
  INS get_ins() { return ins;}
};

// Stack for the Insr structure
static std::unordered_map<ADDRINT, Insr*> insstack;

// This function is called before every instruction is executed
VOID protect(uint64_t addr)
{
  if (addr > 0x700000000000)
        return;
    if (!key)
        return;
  // Initialize the diassembled instruction
  string insdis = insstack[addr]->get_insDis();
  INS ins = insstack[addr]->get_ins();
    if (INS_OperandCount(ins) > 0)
    {
        if (REG(INS_OperandReg(ins, 0)) == REG_AH)
            std::cout << "register: " << REG_StringShort(REG(INS_OperandReg(ins, 0))) << '\n';
    }

  if((INS_Opcode(ins) == XED_ICLASS_ADD || INS_Opcode(ins) == XED_ICLASS_SUB) &&
   INS_OperandIsImmediate(ins, 1))
    {
      int value = INS_OperandImmediate(ins, 1);
        std::cout << "value: " << dec<<value << '\n';
    }
  std::cout << hex <<addr << "\t" << insdis << std::endl;
}

// Pin calls this function every time a new instruction is encountered
VOID Instruction(INS ins, VOID *v)
{
        if (INS_Address(ins) > 0x700000000000)
        return;

    insstack.insert(std::make_pair(INS_Address(ins), new Insr(string(INS_Disassemble(ins)),
    ins)));
    // if (REG_valid_for_iarg_reg_value(INS_MemoryIndexReg(ins)))
    //   std::cout << "true" << '\n';
    // Insert a call to docount before every instruction, no arguments are passed
    INS_InsertCall(ins, IPOINT_BEFORE, (AFUNPTR)protect, IARG_ADDRINT, INS_Address(ins),
  IARG_END);
}

// Lock Routine
void mutex_lock()
{
key = 0;
std::cout<<"out\n";
}
void mutex_unlock()
{
    key = 1;
    std::cout<<"in\n";
}

void Routine(RTN rtn, VOID *V)
{
    if (RTN_Name(rtn) == "main")
    {
        RTN_Open(rtn);
        RTN_InsertCall(rtn, IPOINT_BEFORE, (AFUNPTR)mutex_unlock, IARG_END);
        RTN_InsertCall(rtn, IPOINT_AFTER, (AFUNPTR)mutex_lock, IARG_END);
        RTN_Close(rtn);
    }
}

INT32 Usage()
{
    cerr << "This tool counts the number of dynamic instructions executed" << endl;
    cerr << endl << KNOB_BASE::StringKnobSummary() << endl;
    return -1;
}

int main(int argc, char * argv[])
{
    // Initialize the symbol table
    PIN_InitSymbols();

    // Initialize pin
    if (PIN_Init(argc, argv)) return Usage();

    PIN_SetSyntaxIntel();

    // Routine instrumentation
    RTN_AddInstrumentFunction(Routine, 0);

    // Register Instruction to be called to instrument instructions
    INS_AddInstrumentFunction(Instruction, 0);

    // Start the program, never returns
    PIN_StartProgram();

    return 0;
}

I have few questions regarding that.

How can I understand such a behavior? And how can I detect rsp if I want to? Lastly, how does the instruction prints rsp, but REG(INS_OperandReg(ins, 0)) == REG_STACK_PTR can not detect it?

Solution

  • The INS objects are only valid inside instrumentation routines, such as your Instruction routine. The INS type is nothing but a 32-bit integer that identifies an instruction. The Pin runtime internally maintains a table that maps these 32-bit integers to specific static instructions. It creates such a table whenever it's about to call an instrumentation routine. When the instrumentation routine returns, there is no guarantee that any of these identifiers map to the same static instructions and they may not even be valid. So when you save a copy of an INS object in the following line of code:

    insstack.insert(std::make_pair(INS_Address(ins), new Insr(string(INS_Disassemble(ins)),
    ins)));

    that copy is only useful in the same instance of the Instruction routine. The next time the Instruction routine is called (or any other instrumentation routine), an instruction identifier might be reused for other instructions.

    If you really want to pass an instruction to an analysis routine, you have two options: