Linux 获取 x86 的当前指令地址
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Get address of current instruction for x86
提问by MetallicPriest
I am using Linux with x86 (64 bit to be precise). Is there a way I can get the address of the current instruction. Actually I want to write my own simplified versions of setjmp/longjmp. Here, R.. posted a simplified version of longjmp. Any idea how setjmpis implemented. A simplified version that is, without taking into account of exceptions and signals etc...
我在 x86（准确地说是 64 位）上使用 Linux。有没有办法获得当前指令的地址。其实我想编写我自己的setjmp/longjmp的简化版本。在这里，R.. 发布了longjmp的简化版本。任何想法setjmp是如何实现的。一个简化版本，即不考虑异常和信号等...
采纳答案by NPE
I believe in 64-bit code you can simply do lea rax, [rip].
我相信 64 位代码你可以简单地做lea rax, [rip]。
The 32-bit idiom is:
32 位的习惯用法是：
call next
next: pop eax
回答by MetallicPriest
Thissite gives a simple version of setjmp and longjmp, which is as follows.
该网站提供了一个简单的版本setjmp和longjmp的，这是如下。
#include "setjmp.h"
#define OFS_EBP 0
#define OFS_EBX 4
#define OFS_EDI 8
#define OFS_ESI 12
#define OFS_ESP 16
#define OFS_EIP 20
__declspec(naked) int setjmp(jmp_buf env)
{
__asm
{
mov edx, 4[esp] // Get jmp_buf pointer
mov eax, [esp] // Save EIP
mov OFS_EIP[edx], eax
mov OFS_EBP[edx], ebp // Save EBP, EBX, EDI, ESI, and ESP
mov OFS_EBX[edx], ebx
mov OFS_EDI[edx], edi
mov OFS_ESI[edx], esi
mov OFS_ESP[edx], esp
xor eax, eax // Return 0
ret
}
}
__declspec(naked) void longjmp(jmp_buf env, int value)
{
__asm
{
mov edx, 4[esp] // Get jmp_buf pointer
mov eax, 8[esp] // Get return value (eax)
mov esp, OFS_ESP[edx] // Switch to new stack position
mov ebx, OFS_EIP[edx] // Get new EIP value and set as return address
mov [esp], ebx
mov ebp, OFS_EBP[edx] // Restore EBP, EBX, EDI, and ESI
mov ebx, OFS_EBX[edx]
mov edi, OFS_EDI[edx]
mov esi, OFS_ESI[edx]
ret
}
}
回答by BlueRaja - Danny Pflughoeft
The offset-into-the-current-segment register (EIP) is not normally accessible. However, there is a hackish-way to read it indirectly - you trick the program into pushing the value of EIP onto the stack, then just read it off. You could create a subroutine that looks like this:
偏移到当前段寄存器 ( EIP) 通常不可访问。然而，有一种间接读取它的hackish-way - 你欺骗程序将EIP的值推送到堆栈上，然后读取它。您可以创建一个如下所示的子程序：
GetAddress:
mov eax, [esp]
ret
...
call GetAddress ; address of this line stored in eax
Or, even simpler:
或者，更简单：
call NextLine
NextLine:
pop eax ; address of previous line stored in EAX
If you use a CALL FARinstruction, the segment value (CS) will be pushed on the stack as well.
如果使用CALL FAR指令，段值 ( CS) 也将被压入堆栈。
If you're using C, there are various compiler-specific C-extensions you could use on this page. See also this interesting article.
回答by Basile Starynkevitch
If using GCC, you could also use __builtin_return_address
如果使用 GCC，您还可以使用 __builtin_return_address
