RE: Getting a process' EIP address (and other registers)

["Hua Zhong" <hzhong@cisco.com>]

The EIP is not on the user space stack. It's a system call, not a function
call, and the EIP is where the system call is made.

Upon entering kernel all registeres are saved on the kernel stack. You can
get it by the following:

struct pt_regs *regs = *(((struct pt_regs *)(THREAD_SIZE + (unsigned
long)current)) - 1);

> 	I have written a 2.4 kernel module that is triggered upon an
> IOCTL from a user application. Once triggered the kernel module dumps
> out the contents of the system memory and the x86 CPU 
> architecture state
> to separate files.
> 	I'm writing to the list because my method for getting registers
> from the user process before it entered the kernel is producing
> incorrect results. I've looked all through the kernel code 
> and searched
> all over the web for an answer to this specific question and found
> nothing relevant.
> 	In my research how to do this I've found that the stack pointer
> for the user process before it entered the kernel is stored in
> current->thread.esp0. From there the registers for the user 
> process are
> stored at offsets from that location (for example, EIP is 
> supposed to be
> 0x28 unsigned char bytes from esp0). I have code to get the EIP as
> follows:
> 
> unsigned char *stack_address, *eip;
> 
> stack_address = (unsigned char *)(current->thread.esp0);
> eip = stack_address + 0x28;
> printk("eip = %08lx\n", *(unsigned long *)eip);
> 
> Like I mentioned this is producing incorrect results. How do I access
> the user process' general purpose registers contents as they 
> were before
> the kernel was entered (or correct my code above if I'm accessing
> something wrong)?
> 	Thanks in advance for any help offered.
> -
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