我正在尝试从内核模块调用系统调用,我有以下代码:
set_fs( get_ds() ); // lets our module do the system-calls
// Save everything before systemcalling
asm (" push %rax ");
asm (" push %rdi ");
asm (" push %rcx ");
asm (" push %rsi ");
asm (" push %rdx ");
asm (" push %r10 ");
asm (" push %r8 ");
asm (" push %r9 ");
asm (" push %r11 ");
asm (" push %r12 ");
asm (" push %r15 ");
asm (" push %rbp ");
asm (" push %rbx ");
// Invoke the long sys_mknod(const char __user *filename, int mode, unsigned dev);
asm volatile (" movq $133, %rax "); // system call number
asm volatile (" lea path(%rip), %rdi "); // path is char path[] = ".."
asm volatile (" movq mode, %rsi "); // mode is S_IFCHR | ...
asm volatile (" movq dev, %rdx "); // dev is 70 >> 8
asm volatile (" syscall ");
// POP EVERYTHING
asm (" pop %rbx ");
asm (" pop %rbp ");
asm (" pop %r15 ");
asm (" pop %r12 ");
asm (" pop %r11 ");
asm (" pop %r9 ");
asm (" pop %r8 ");
asm (" pop %r10 ");
asm (" pop %rdx ");
asm (" pop %rsi ");
asm (" pop %rcx ");
asm (" pop %rdi ");
asm (" pop %rax ");
set_fs( savedFS ); // restore the former address-limit value 这段代码无法工作,导致系统崩溃(这是一个内核模块)。
带有重定位信息的那段代码的转储是:
2c: 50 push %rax
2d: 57 push %rdi
2e: 51 push %rcx
2f: 56 push %rsi
30: 52 push %rdx
31: 41 52 push %r10
33: 41 50 push %r8
35: 41 51 push %r9
37: 41 53 push %r11
39: 41 54 push %r12
3b: 41 57 push %r15
3d: 55 push %rbp
3e: 53 push %rbx
3f: 48 c7 c0 85 00 00 00 mov $0x85,%rax
46: 48 8d 3d 00 00 00 00 lea 0x0(%rip),%rdi # 4d <init_module+0x4d>
49: R_X86_64_PC32 path-0x4
4d: 48 83 c7 04 add $0x4,%rdi
51: 48 8b 34 25 00 00 00 mov 0x0,%rsi
58: 00
55: R_X86_64_32S mode
59: 48 8b 14 25 00 00 00 mov 0x0,%rdx
60: 00
5d: R_X86_64_32S dev
61: 0f 05 syscall
63: 5b pop %rbx
64: 5d pop %rbp
65: 41 5f pop %r15
67: 41 5c pop %r12
69: 41 5b pop %r11
6b: 41 59 pop %r9
6d: 41 58 pop %r8
6f: 41 5a pop %r10
71: 5a pop %rdx
72: 5e pop %rsi
73: 59 pop %rcx
74: 5f pop %rdi
75: 58 pop %rax 我想知道..为什么49: R_X86_64_PC32路径-0x4中有-0x4偏移量?
我的意思是:模式和开发应该自动解决,没有问题,但路径呢?为什么偏移量为-0x4?
我试着用“补偿”
lea 0x0(%rip),%rdi //这以某种方式添加了-0x4偏移量加上$0x4,%rdi ....
但是代码还是崩溃了。
我哪里错了?
发布于 2012-03-19 13:39:00
关于这里发生的事情,我猜测是堆栈问题。与int $0x80不同,syscall指令不会为内核设置堆栈。如果您查看来自system_call:的实际代码,您将看到类似于SWAPGS_UNSAFE_STACK的内容。该宏的核心是SwapGS指令-参见第152页here。当进入内核模式时,内核需要一种方法来拉出指向其数据结构的指针,而这条指令正是让它这样做的。为此,它将用户%gs寄存器交换为特定于模型的寄存器中保存的值,然后可以从该寄存器中拉出内核模式堆栈。
您可以想象,一旦调用了syscall入口点,这个交换就会生成错误的值,因为您已经处于内核模式,并且内核开始尝试使用一个伪造的堆栈。您可以尝试手动调用SwapGS,使内核的SwapGS得到它所期望的结果,然后看看是否能正常工作。
发布于 2012-03-19 03:48:17
看起来你不能用这种方式做那件事。请参阅system_call之前的comment
/*
* Register setup:
* rax system call number
* rdi arg0
* rcx return address for syscall/sysret, C arg3
* rsi arg1
* rdx arg2
* r10 arg3 (--> moved to rcx for C)
* r8 arg4
* r9 arg5
* r11 eflags for syscall/sysret, temporary for C
* r12-r15,rbp,rbx saved by C code, not touched.
*
* Interrupts are off on entry.
* Only called from user space.
*
* XXX if we had a free scratch register we could save the RSP into the stack frame
* and report it properly in ps. Unfortunately we haven't.
*
* When user can change the frames always force IRET. That is because
* it deals with uncanonical addresses better. SYSRET has trouble
* with them due to bugs in both AMD and Intel CPUs.
*/因此,您不能从内核调用syscall。但您可以尝试使用int $0x80来实现此目的。在我看来,kernel_execve存根使用该trick
https://stackoverflow.com/questions/9758869
复制相似问题