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自修改代码中的可能指令缓存同步问题?

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许多相关问题<How is x86 instruction cache synchronized?>提及x86应该在自修改代码中正确处理i-cache同步 . 我编写了下面这段代码,它可以在与执行交错的不同线程上打开和关闭函数调用 . 我使用比较和交换操作作为额外的防护,以便修改是原子的 . 但我得到间歇性崩溃(SIGSEGV,SIGILL)并分析核心转储使我怀疑处理器是否正在尝试执行部分更新的指令 . 下面给出的代码和分析 . 可能是我在这里遗漏了一些东西 . 如果是这样,请告诉我 .

toggle.c

#include <stdio.h>
#include <inttypes.h>
#include <time.h>
#include <pthread.h>
#include <sys/mman.h>
#include <errno.h>
#include <unistd.h>

int active = 1; // Whether the function is toggled on or off
uint8_t* funcAddr = 0; // Address where function call happens which we need to toggle on/off
uint64_t activeSequence = 0; // Byte sequence for toggling on the function CALL
uint64_t deactiveSequence = 0; // NOP byte sequence for toggling off the function CALL

inline int modify_page_permissions(uint8_t* addr) {

  long page_size = sysconf(_SC_PAGESIZE);
  int code = mprotect((void*)(addr - (((uint64_t)addr)%page_size)), page_size,
    PROT_READ | PROT_WRITE | PROT_EXEC);

  if (code) {
    fprintf(stderr, "mprotect was not successfull! code %d\n", code);
    fprintf(stderr, "errno value is : %d\n", errno);
    return 0;
  }

  // If the 8 bytes we need to modify straddles a page boundary make the next page writable too
  if (page_size - ((uint64_t)addr)%page_size < 8) {
    code = mprotect((void*)(addr-((uint64_t)addr)%page_size+ page_size) , page_size,
      PROT_READ | PROT_WRITE | PROT_EXEC);
    if (code) {
      fprintf(stderr, "mprotect was not successfull! code %d\n", code);
      fprintf(stderr, "errno value is : %d\n", errno);
      return 0;;
    }
  }

  return 1;
}

void* add_call(void* param) {

  struct timespec ts;
  ts.tv_sec = 0;
  ts.tv_nsec = 50000;

  while (1) {
    if (!active) {
      if (activeSequence != 0) {
        int status = modify_page_permissions(funcAddr);
        if (!status) {
          return 0;
        }

        uint8_t* start_addr = funcAddr - 8;

        fprintf(stderr, "Activating foo..\n");
        uint64_t res = __sync_val_compare_and_swap((uint64_t*) start_addr,
                                    *((uint64_t*)start_addr), activeSequence);
        active = 1;
      } else {
        fprintf(stderr, "Active sequence not initialized..\n");
      }
    }

    nanosleep(&ts, NULL);
  }

}

int remove_call(uint8_t* addr) {

  if (active) {
    // Remove gets called first before add so we initialize active and deactive state byte sequences during the first call the remove
    if (deactiveSequence == 0) {
      uint64_t sequence =  *((uint64_t*)(addr-8));
      uint64_t mask = 0x0000000000FFFFFF;
      uint64_t deactive = (uint64_t) (sequence & mask);
      mask = 0x9090909090000000; // We NOP 5 bytes of CALL instruction and leave rest of the 3 bytes as it is

      activeSequence = sequence;
      deactiveSequence = deactive |  mask;
      funcAddr = addr;
    }

    int status = modify_page_permissions(addr);
    if (!status) {
      return -1;
    }

    uint8_t* start_addr = addr - 8;

    fprintf(stderr, "Deactivating foo..\n");
    uint64_t res = __sync_val_compare_and_swap((uint64_t*)start_addr,
                                  *((uint64_t*)start_addr), deactiveSequence);
    active = 0;
    // fprintf(stderr, "Result : %p\n", res);
  }
}

int counter = 0;

void foo(int i) {

  // Use the return address to determine where we need to patch foo CALL instruction (5 bytes)
  uint64_t* addr = (uint64_t*)__builtin_extract_return_addr(__builtin_return_address(0));

  fprintf(stderr, "Foo counter : %d\n", counter++);
  remove_call((uint8_t*)addr);
}

// This thread periodically checks if the method is inactive and if so reactivates it
void spawn_add_call_thread() {
  pthread_t tid;
  pthread_create(&tid, NULL, add_call, (void*)NULL);
}

int main() {

  spawn_add_call_thread();

  int i=0;
  for (i=0; i<1000000; i++) {
    // fprintf(stderr, "i : %d..\n", i);
   foo(i);
  }

  fprintf(stderr, "Final count : %d..\n\n\n", counter);
}

Core dump analysis

Program terminated with signal 4, Illegal instruction.
#0  0x0000000000400a28 in main () at toggle.c:123
(gdb) info frame
 Stack level 0, frame at 0x7fff7c8ee360:
   rip = 0x400a28 in main (toggle.c:123); saved rip 0x310521ed5d
 source language c.
 Arglist at 0x7fff7c8ee350, args:
 Locals at 0x7fff7c8ee350, Previous frame's sp is 0x7fff7c8ee360
 Saved registers:
 rbp at 0x7fff7c8ee350, rip at 0x7fff7c8ee358
(gdb) disas /r 0x400a28,+30
 Dump of assembler code from 0x400a28 to 0x400a46:
  => 0x0000000000400a28 <main+64>:   ff (bad)
     0x0000000000400a29 <main+65>:   ff (bad)
     0x0000000000400a2a <main+66>:   ff eb  ljmpq  *<internal disassembler error>
     0x0000000000400a2c <main+68>:   e7 48  out    %eax,$0x48
 (gdb) disas /r main
  Dump of assembler code for function main:
     0x00000000004009e8 <+0>:    55 push   %rbp
     ...
     0x0000000000400a24 <+60>:   89 c7  mov    %eax,%edi
     0x0000000000400a26 <+62>:   e8 11 ff ff ff callq  0x40093c <foo>
     0x0000000000400a2b <+67>:   eb e7  jmp    0x400a14 <main+44>

因此可以看出,指令指针似乎位于CALL指令内的地址内,并且处理器显然试图执行该未对齐的指令,从而导致非法指令错误 .

2 回答

  • 3

    我认为你的问题是你用5个1字节的NOP替换了一个5字节的CALL指令 . 考虑当你的线程执行3个NOP时会发生什么,然后你的主线程决定重新交换CALL指令 . 你的线程的PC将在CALL指令的中间三个字节,因此会执行一个意外的并且可能是非法的指令 .

    您需要做的是将5字节CALL指令与5字节NOP交换 . 你只需要找到一个什么都不做的多字节指令(例如或者对自己进行寄存器),如果你需要一些额外的字节,可以在前面添加一些前缀字节,例如gs覆盖前缀和地址大小覆盖前缀(两者都是哪个什么都不做) . 通过使用5字节的NOP,您的线程将保证在CALL指令处或通过CALL指令,但从不在其内部 .

  • 3

    在80x86上,大多数调用使用相对位移,而不是绝对地址 . 本质上它是“在这里调用代码<displacement>”而不是“在<address>调用代码” .

    对于64位代码,位移可以是8位或32位 . 它永远不会是64位 .

    例如,对于一个2字节的"call with 8-bit displacement"指令,您将在调用指令之前丢弃6个字节, call 操作码本身和指令的操作数(位移) .

    再举一个例子,对于一个5字节的"call with 32-bit displacement"指令,你将在调用指令之前删除3个字节, call 操作码本身和指令的操作数(位移) .

    然而...

    这些't the only way to call. For example, you can call using a function pointer, where the address of the code being called is not in the instruction at all (but may be in a register or be a variable in memory). There'也是一个名为"tail call optimisation"的优化,其中 call 后跟 retjmp 替换(可能还有一些额外的堆栈用于传递参数,清理调用者的局部变量等) .

    实质上;你的代码严重破坏,你无法涵盖所有可能的极端情况,你不应该这样开始,你可能应该使用函数指针而不是自修改代码(这将更快更容易便携式) .

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