Re: [rfc] 16 bit real-mode for the i386

[Andrew Cagney <ac131313@ges.redhat.com>]

> Now, my i386 knowledge is a bit rusty.  But:
> 
> 
>> +static CORE_ADDR
>> +i386_read_pc (ptid_t ptid)
>> +{
>> +  CORE_ADDR pc = read_register_pid (PC_REGNUM, ptid);
>> +  if (i386_real_mode_p)
>> +    {
>> +      CORE_ADDR cs = read_register_pid (CS_REGNUM, ptid);
>> +      pc = pc + (cs << 4);
>> +    }
>> +  return pc;
>> +}
>> +
>> +static void
>> +i386_write_pc (CORE_ADDR pc, ptid_t ptid)
>> +{
>> +  if (i386_real_mode_p)
>> +    {
>> +      CORE_ADDR cs = read_register_pid (CS_REGNUM, ptid);
>> +      pc = pc - (cs << 4);
>> +    }
>> +  write_register_pid (PC_REGNUM, pc, ptid);
>> +}
>> +
> 
> 
> Left shift of _four_?  Surely the PC is more than four bits.

It is.  From an ia32 manual (www.intel.com):

16.1.1.    Address Translation in Real-Address Mode
In real-address mode, the processor does not interpret segment selectors 
as indexes into a descriptor table; instead, it uses them directly to 
form linear addresses as the 8086 processor does. It shifts the segment 
selector left by 4 bits to form a 20-bit base address (see Figure 16-1).
The offset into a segment is added to the base address to create a 
linear address that maps directly
to the physical address space.

Andrew