Re: how are debug registers supposed to work?

[Ben Johnson <ben@blarg.net>]

I found the problem.  The addresses I'm attempting to use are logical
addresses, not linear.  The (2.0) kernel data segment's base address is
0xc0000000, so in order to get a linear address I have to add that base
address to it.

altered code that's now trapping in the right place:


schedule()
{
...
	static unsigned long has_run = 1;
	static unsigned long has_run_2 = 0;

        if( ! has_run && jiffies > 7000 )
        {
                has_run = 1;
                has_run_2 = 0;

                /* setup the debug registers */
                asm ("movl %%cr4, %%edx\n"   /* debug extensions */
                     "  orl $0x8, %%edx\n"
                     "  movl %%edx, %%cr4\n"
                     "  movl %0, %%db0\n"    /* push into db regs */
                     "  movl %1, %%db7\n"
                     "  lgdt 0x00106852\n"   /* pentium may need this */
                     : /* no output */
                     :"a"(0xc0000000 + ((unsigned long)&has_run_2)),
                      "b"(0x000f2202)
                      /*"m"((((char *)&gdt)-6))*/
                     :"%edx"
                    );
        }

        if( has_run && ! has_run_2 )  /* debug reg generate exception */
        {
		/* whatever */
		has_run_2 = 0xffffffff
        }
...
}


I'm sure the debug extensions aren't needed.  I put in the lgdt
instruction because I read section in 18.17.4 of Intel's Software
Development Manual (Volume 3) that it may help Pentium processors
recognize breakpoints.  no other processors need that though.

- Ben