Memory access error sys_rt_sigaction (signal handler)

泄露秘密 提交于 2019-11-29 11:45:04

In x86-64 linux, it's mandatory to supply a sa_restorer and you haven't done so.

The relevant part of kernel source:

            /* x86-64 should always use SA_RESTORER. */
            if (ksig->ka.sa.sa_flags & SA_RESTORER) {
                    put_user_ex(ksig->ka.sa.sa_restorer, &frame->pretcode);
            } else {
                    /* could use a vstub here */
                    err |= -EFAULT;
            }

The C library wrapper does this for you:

  kact.sa_flags = act->sa_flags | SA_RESTORER;

  kact.sa_restorer = &restore_rt;

With the updated code you do indeed have a restorer, but you have two problems: it's broken and you pass it wrong. Looking at the above mentioned C library source you can find this comment:

/* The difference here is that the sigaction structure used in the
   kernel is not the same as we use in the libc.  Therefore we must
   translate it here.  */

Also, you can't have a C++ function as restorer due to the function prologue. Furthermore, calling printf from a signal handler is not supported (but works here). Finally, as David Wohlferd pointed out, your clobbers are wrong. All in all, the following could be a reworked version:

#include<stdio.h>
#include<unistd.h>
#include<time.h>

void handler(int){
    const char msg[] = "handler\n";
    write(0, msg, sizeof(msg));
}

extern "C" void restorer();
asm volatile("restorer:mov $15,%rax\nsyscall");

struct kernel_sigaction {
        void (*k_sa_handler) (int);
        unsigned long sa_flags;
        void (*sa_restorer) (void);
        unsigned long sa_mask;
};

struct kernel_sigaction act{handler};
timespec ts{10,0};

int main(){
 act.sa_flags=0x04000000;
 act.sa_restorer=&restorer;

 asm volatile("\
 mov $13,%%rax\n\
 mov %0,%%rdi\n\
 mov %1,%%rsi\n\
 mov %2,%%rdx\n\
 mov $8,%%r10\n\
 syscall\n\
 "::"i"(7),"p"(&act),"p"(0):"rax","rcx", "rdi","rsi","rdx","r8", "r9", "r10", "r11");

 nanosleep(&ts,0);
}

It's still hacky, and you shouldn't really be doing it this way, obviously.

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