As we know the BIOS Interrupt (INT) 0x19 which searches for a boot signature (0xAA55). Loads and executes our bootloader at 0x7C00 if it found.
My Question : Why 0x7C00? What is the reason ? How to evaluate it through some methods?
Asked
Active
Viewed 2,262 times
8
-
I don't think anyone here wrote the BIOS, so we couldn't tell you. Do you have a specific problem? – John Saunders Mar 26 '13 at 01:23
-
9About 1 minute on Google produces this badly-formatted and grammatically poor (but somewhat promising looking) result: http://www.glamenv-septzen.net/en/view/6 – Cody Gray - on strike Mar 26 '13 at 03:14
3 Answers
3
Maybe because the MBR is loaded into the memory (by the BIOS) into the 0x7c00 address then int 0x19 searches for the MBR sector signature 0xAA55 on sector 0x7c00
about 0xAA55:
Not a checksum, but more of a signature. It does provide some simple evidence that some MBR is present.
0xAA55 is also an alternating bit pattern: 1010101001010101
It's often used to help determine if you are on a little-endian or big-endian system, because it will read as either AA55 or 55AA. I suspect that is part of why it is put on the end of the MBR.
about 0x7c00:
Check this website out (this might help u in finding the answer): https://www.glamenv-septzen.net/en/view/6
MMG
- 3,226
- 5
- 16
- 43
1
( It seems like this duplicates following questions:
What is significance of memory at 0000:7c00 to booting sequence?
Does the BIOS copy the 512-byte bootloader to 0x7c00 )
Inspired by an answer to the former, I would quote two sources:
So, first answer the question about the 16KB model from David Bradley's reply:
It had to boot on a 32KB machine. DOS 1.0 required a minimum of 32KB, so we weren't concerned about attempting a boot in 16KB.
To execute DOS 1.0, at least 32KB is required, so the 16KB model has not been considered.
Followed by the answer to "Why 32KB-1024B?":
We wanted to leave as much room as possible for the OS to load itself within the 32KB. The 808x Intel architecture used up the first portion of the memory range for software interrupts, and the BIOS data area was after it. So we put the bootstrap load at 0x7C00 (32KB-1KB) to leave all the room in between for the OS to load. The boot sector was 512 bytes, and when it executes it'll need some room for data and a stack, so that's the other 512 bytes . So the memory map looks like this after INT 19H executes:
No, that case was out of consideration. One of IBM PC 5150 ROM BIOS Developer Team Members, Dr. David Bradley says:
"DOS 1.0 required a minimum of 32KB, so we weren't concerned about attempting a boot in 16KB."
(Note: DOS 1.0 required 16KiB minimum ? or 32KiB ? I couldn't find out which correct. But, at least, in 1981's early BIOS development, they supposed that 32KiB is DOS minimum requirements.)
BIOS developer team decided 0x7C00 because:
- They wanted to leave as much room as possible for the OS to load itself within the 32KiB.
- 8086/8088 used 0x0 - 0x3FF for interrupts vector, and BIOS data area was after it.
- The boot sector was 512 bytes, and stack/data area for boot program needed more 512 bytes.
- So, 0x7C00, the last 1024B of 32KiB was chosen.
Once OS loaded and started, boot sector is never used until power reset. So, OS and application can use the last 1024B of 32KiB freely.
I hope this answer is based enough to be sure why / how it happened so.
saulius2
- 249
- 4
- 8
0
This probably dead but I'm going to answer.
At the start of any bootloader when you set the origin of the segment to 0x7c00 then the registers jump address to that as well. So ideally if you check out some online resources that tell you how to use the int 0x19 command they will guide you on how to jump to another address.
To fix this you would ideally, reset the stack to 0 at the start of every jump to an new address.
