Question

作为x86指令的解决方法，我使用了一个看起来像.byte 0xf1, 0xc1的操作码，并试图在gdb中执行该操作码。该行的反汇编显示

f1      icebp

因此，当获取F1的第一个字节时，它将被识别为称为icebp的指令。已知这是未记录的说明。我在SDM中唯一发现的是INT部分的脚注

The mnemonic ICEBP has also been used for the instruction with opcode F1

继续说gdb

Cannot access memory at address 0x1ffffc20

那那个地址是什么？它是如何产生的？是物理的还是虚拟的？以及如何测试其实际功能？

更新：

GDB操作如下所示：

(gdb) list
1       void main()
2       {
3         __asm__(".byte 0xf1, 0xc1");
4       }
(gdb) set disassembly-flavor intel
(gdb) disass /r main
Dump of assembler code for function main:
   0x00000000004004ed <+0>:     55      push   rbp
   0x00000000004004ee <+1>:     48 89 e5        mov    rbp,rsp
   0x00000000004004f1 <+4>:     f1      icebp
   0x00000000004004f2 <+5>:     c1 5d c3 66     rcr    DWORD PTR [rbp-0x3d],0x66
End of assembler dump.
(gdb) b main
Breakpoint 1 at 0x4004f3: file machine2.c, line 4.
(gdb) run
Starting program: /home/mahmood/Documents/./machine2

Program received signal SIGTRAP, Trace/breakpoint trap.
0x00000000004004f2 in main () at machine2.c:2
2       {
Missing separate debuginfos, use: debuginfo-install glibc-2.17-196.el7.x86_64
(gdb) x/i $pc
=> 0x4004f2 <main+5>:   rcr    DWORD PTR [rbp-0x3d],0x66
(gdb) n
Cannot access memory at address 0x1ffffc20
(gdb) x/i $pc
=> 0x4004f5:    nop    WORD PTR cs:[rax+rax*1+0x0]

UDPATE2：

删除c1后，调试器将无法在asm行设置断点。

(gdb) list
1       void main()
2       {
3         __asm__(".byte 0xf1");
4       }
(gdb) b machine2.c:3
Breakpoint 1 at 0x4004f2: file machine2.c, line 3.
(gdb) run
Starting program: /home/mahmood/Documents/./machine2

Breakpoint 1, main () at machine2.c:4
4       }
Missing separate debuginfos, use: debuginfo-install glibc-2.17-196.el7.x86_64

ICEBP组装说明

0 个答案: