| ??? 09/08/05 15:16 Read: times |
#100820 - irrelevent: irrelevent? Responding to: ???'s previous message |
Andy Neil said:
Memory is memory - it is addressed in the normal way. It's not the memory itself that is segmented - it's just the way the x86 works with it internally. exactly thats why i called it x86 memory The x86 is a 16-bit processor, so all its internal registers & data paths are 16 bits wide. But the x86 has a 20-bit address space; the Segment:Offset business is just the way it manages to get 20 bits out of two 16-bit registers! what is proved? All the memory ever sees is a 20-bit address - it knows nothing about segmentation! ?? processor is executing the code so rollover ocuurs in processor! The x86 has a whole bunch of registers that can be used as Segment registers, and plenty of others that can be used as offsets... ok but here op talks about the IP roll over all those register (SS, DS ES (FS GS)) cant be used to execute code, the instruction fetch always occurs corresponding to the CS. On execution, procvessor incremets IP. if rollover occurs all it will start executing the CS:0000 i.e. same segment offset zero. it will not jump to 0000:0000, that was my question. Abhishek |
| Topic | Author | Date |
| Micro Controller powered? | 01/01/70 00:00 | |
| bible study time? | 01/01/70 00:00 | |
| Reboot ? | 01/01/70 00:00 | |
| See reply (Off-Topic) | 01/01/70 00:00 | |
| Not A PC | 01/01/70 00:00 | |
| Reboot doubt | 01/01/70 00:00 | |
| Off Topic. | 01/01/70 00:00 | |
| Rampant code execution | 01/01/70 00:00 | |
| CS:IP | 01/01/70 00:00 | |
| Irrelevant? | 01/01/70 00:00 | |
| irrelevent: irrelevent? | 01/01/70 00:00 | |
| TOTALLY irrelevant, at least at a '51 fo | 01/01/70 00:00 | |
| Already established! | 01/01/70 00:00 | |
| Wrong | 01/01/70 00:00 | |
| You Sure? | 01/01/70 00:00 | |
If you want a proper answer... | 01/01/70 00:00 |