One day, I was curious about how the computer system goes from booting to actually loading up an operating system. Obviously, it must retrieve the operating system from disk at some point, so I decided to investigate this. The first step in this process is reading the MBR, or Master Boot Record of the hard drive. The MBR is used to store data about where the OS is stored on the drive.

I figured the MBR would be interesting to learn a little bit more about, so I decided to load it up into IDA Pro, a tool for disassembling programs, and see what I could find out.



I learned a lot and had a lot of fun, so I’m presenting it here to share my results.

For this analysis, I assume that you are familiar with the x86 architecture and assembly. If not, WikiBooks has some great information about it here. I am also using IDA Pro to do this. I have tried to provide comments and labeling in my IDA work, but I also explain each block of code separately.

The first step in figuring out the MBR was to actually get a copy of it I could work with. To do this, I used Hex Workshop, which offers a way to do a binary copy of hard disks. The MBR is always located as the first sector of the disk, so I used Hex Workshop and extracted this file. It is only 1 sector long, so it is a mere 512 bytes.

After looking at Wikipedia, I figured out the MBR structure was:

0000h - 01B7h Code Area (440 bytes) 01B8h - 01BBh Disk Signature (4 bytes) 01BCh - 01BDh Generally Zeroed out (2 bytes) 01BEh - 01FDh List of Partition Records (4 16-byte structures) 01FEh - 01FFh MBR Signature (2 bytes - Must be AA55h)

The above structure is what gets loaded into memory and executed. The code area is going to do a few different things, which I look at below. The disk signature can be used to uniquely identify a hard disk. The partition records define different operating systems and partitions on the hard disk. Have you ever noticed how you have to jump through some hoops if you want to have more than 4 operating systems or partitions on your computer? Well, that’s because you only have 4 partition records available. The MBR signature helps illustrate that this is in fact an MBR structure.

The partition records are a pretty important part of the MBR, so I also examined their structure. It is:

0000h - 0000h Status byte (80h for bootable, 00h for non-bootable, others are invalid) (1 byte) 0001h - 0003h CHS address of first absolute sector (3 bytes) 0004h - 0004h Partition type (1 byte) 0005h - 0007h CHS address of last absolute sector (3 bytes) 0008h - 000Bh LBA of first absolute sector (4 bytes) 000Ch - 000Fh Number of sectors in partition (4 bytes)

That is a pretty simple structure; just a start and stop address, length, and a few informational bytes. The CHS format is a little confusing though. CHS stands for Cylinder-Head-Sector and defines an address inside of a physical hard drive. You can read more about CHS here. The LBA address stands for Logical Block Address. LBA is a format to linearly address space on the hard drive, rather than defining 3 separate numbers for Cylinder-Head-Sector format. More information about LBA is here.

After examining the file MBR structure, I loaded the binary file into IDA Pro. Since it is a binary file, IDA doesn’t know where the correct segments or entry points are, or even if it is 16 or 32 bit code. Since we don’t have an OS yet, we know that this code is 16 bit. Looking at the MBR structure, the code starts at the very first byte of the file. Knowing this, I configured IDA and converted the first few bytes to code. I got the following:

s0:0000 ; --------------------------------------------------------------------------- s0:0000 s0:0000 ; Segment type: Pure code s0:0000 seg000 segment byte public 'CODE' use16 s0:0000 assume cs:seg000 s0:0000 assume es:nothing, ss:nothing, ds:nothing, fs:nothing, gs:nothing s0:0000 xor ax, ax ; Zero out AX s0:0002 mov ss, ax ; Set SS to 0 s0:0004 mov sp, 7C00h ; Set SP to 7C00h s0:0007 mov es, ax ; Set ES to 0 s0:0009 mov ds, ax ; Set DS to 0 s0:000B mov si, 7C00h ; Source for the copy s0:000E mov di, 600h ; This is the destination for the copy s0:0011 mov cx, 200h ; We want to copy 200h bytes, which is the length s0:0011 ; of one sector, i.e. the whole MBR. s0:0014 cld ; Clear Direction Flag s0:0015 rep movsb ; Copies CX bytes from [SI] to [DI] s0:0015 ; i.e. from [7C00h] to [600h] s0:0017 push ax ; New value for CS s0:0018 s0:0018 loc_18: ; New value for IP s0:0018 push 61Ch s0:001B retf ; Pops the top of the stack into IP then pops CS s0:001B ; off next. s0:001B ; i.e. we will run from 0000h:61Ch, s0:001B ; which is where we just copied ourselves to. s0:001B ; s0:001B ; Execution continues at the next instruction.

I have tried to add a lot of comments to show what is going on. Essentially though, what this is doing is copying the MBR from 0000h:7C00h to 0000:600h. This is necessary, because it will later load the OS to 0000h:7C00h, so it needs to get itself out of the way. It does this using the ‘rep movsb’ which does a binary copy of 200 bytes from SI (7C00h) to DI (600h).

An interesting part about the code above is the way that the ‘retf’ instruction is used. This does what is known as a ‘far jump’. This means it not only jumps to a different instruction address, but also a different code segment. Both of these values are popped off of the stack, with the offset being on top and the new segment being second. Before the retf instruction, the program pushes 0 and then ’61Ch’ onto the stack to setup for this far jump. This may seem strange, but since it just copied all the program data to 0000h:0600h, 61Ch is actually the offset to the instruction directly after the retf in the binary.

The next few instructions are:

s0:001C sti ; Enable interrupts s0:001D mov cx, 4 ; This will be used for the loop. s0:001D ; We only want to examine 4 partition records. s0:0020 mov bp, 7BEh ; This is the offset to the first partition s0:0020 ; record. s0:0020 ; In the MBR structure, the first partition s0:0020 ; record is at base+1BEh, s0:0023 so it is 600h + 1BEh = 7BEh s0:0023

First thing this code does is to enable interrupts so that it can be interrupted if necessary. Next, it sets the CX register to 4 and BP to 7BEh. BP is the offset to the partition records (there is a description of these records here). CX is indicating that we will only examine 4 records (since there are only 4 in the MBR). So this is setting up to iterate over the 4 partition records to find the bootable one that we want.

After this is a loop that tries to find a bootable partition entry. The code is:

s0:0023 loc_23: ; CODE XREF: seg000:0030j s0:0023 cmp byte ptr [bp+0], 0 ; Compares the status byte to 0 s0:0027 jl short FoundBootableEntry ; Jumps if the status flag is not 0. s0:0027 ; This will happen if the MSB of [bp+0] s0:0027 ; is 1, essentially saying that if it s0:0027 ; is 0x80, it will jump. s0:0027 ; s0:0027 ; This jump indicates that we have found the s0:0027 ; bootable partition. s0:0029 jnz PrintInvalidPartitionTable ; It's not 0x80 and it's not 0x0, so s0:0029 ; it's invalid. Jump to an error state. s0:002D add bp, 10h ; Go to the next partition record. s0:0030 loop loc_23 ; Loop while CX != 0 s0:0032 int 18h ; TRANSFER TO ROM BASIC s0:0032 ; causes transfer to ROM-based BASIC (IBM-PC) s0:0032 ; often reboots a compatible; often has no effect s0:0034 ; at all

Remember how above we moved the offset of 7BEh into BP? That is so this loop can then examine the partition records. The comparison is checking the status byte of the partition record and comparing it to 0. If it is 0, the first jump will not be taken, nor will the second jump, so 10h is added to BP and the loop is restarted. Advancing the BP register means that we will examine a different partition record. If, after 4 iterations, we have still not found a bootable partition entry, an ‘int 18h’ call will be made. On old IBM PCs, this would run a BASIC interpreter from ROM, but few computers have this. So essentially, an int 18h call will just stop the system. Imagine that, if you have no bootable entries, you’re computer won’t boot!

If the status byte of the parition record was in fact 80h, the first jump would have been taken. The JL instruction does a jump if the status flag is set to 1. This flag will get set by the previous CMP instruction if the high order bit is set in the status byte, i.e. the status byte is 80h. If the entries status byte is neither 80h or 00h, a jump is made to the PrintInvalidPartitionTable location. I’ll talk about that a little bit, but it’s pretty boring (it just prints an error message); when a bootable entry is found, things are much more fun.

The next block of code is run when a bootable entry is found. Here it is:

s0:0034 FoundBootableEntry: ; CODE XREF: seg000:0027j s0:0034 ; seg000:00AEj s0:0034 mov [bp+0], dl ; Save the drive number for later s0:0034 ; Note that since this will most likely be the s0:0034 ; first hard drive, DL will probably be 0x80 s0:0037 push bp s0:0038 mov byte ptr [bp+11h], 5 s0:003C mov byte ptr [bp+10h], 0 ; This is a sentinel value for later s0:0040 s0:0040 loc_40: ; DATA XREF: seg000:014Fr s0:0040 mov ah, 41h ; 'A' s0:0042 mov bx, 55AAh s0:0045 int 13h ; DISK - Installation Check s0:0045 ; CF set on error s0:0045 ; CF cleared on success s0:0045 ; BX = AA55 if installed s0:0045 ; AH = major version of extensions s0:0045 ; CX = API subset s0:0045 ; DH = Extension version s0:0047 pop bp s0:0048 jb short AttemptLoadFromDisk ; Jump if Below (CF=1) s0:004A cmp bx, 0AA55h ; DATA XREF: seg000:0045r s0:004A ; seg000:007Er ... s0:004A ; Compare Two Operands s0:004E jnz short AttemptLoadFromDisk ; Jump if Not Zero (ZF=0) s0:0050 test cx, 1 ; Logical Compare s0:0054 jz short AttemptLoadFromDisk ; Jump if Zero (ZF=1) s0:0056 inc byte ptr [bp+10h] ; This acts like a sentinel value s0:0056 ; for whether or not the INT 13 s0:0056 ; extended read is installed s0:0059

First part of this block does is moves the DL register into where the entry’s status byte used to be. I wasn’t really too sure what was going on here for a long time, since if it overwrites the partition record, won’t it not boot correctly next time? Well, it turns out that the first hard drive is represented by 80h, so most of the time, things will be fine. I don’t have 2 hard drives, so I’m not sure what would happen if you had two hard drives and had your bootable entry on the second hard drive. I suspect that the 2nd hard drive would just have its own MBR and would run that instead. Running code on 1 hard drive and loading data from another hard drive seems a little bit silly anyways, so I’m pretty sure that’s whats going on. If you know more, please leave a comment!

Next, the code saves the partition record to the stack and moves the values 5 and 0 into the status byte. The 5 indicates the number of attempts that will be made to read from disk later. Multiple attempts will be made because the disk read might fail while the disk spins up. The 0 value is a sentinel value for whether or not the BIOS supports the extended interrupt 13h feature. This is an advanced, easier way to load lots of data from disk into memory, but not all BIOSs support it. The code above runs several different checks and if they all succeed, it stores a 1 where it had just stored a 0, indicating the extended read feature is supported. If any of those checks fails, it just jumps down a few lines and continues executing and will use the older method.

The next section of code loads the first sector of the OS into memory, using a different method depending on whether or not the extended read is supported. Here it is:

s0:0059 AttemptLoadFromDisk: ; CODE XREF: seg000:0048j s0:0059 ; seg000:004Ej ... s0:0059 pushad ; Save all our registers for a little bit s0:005B cmp byte ptr [bp+10h], 0 ; Did our sentinel value get changed? s0:005F jz short InstallationFailed ; DATA XREF: seg000:0032r s0:005F ; Jump if Zero (ZF=1) s0:0061 push large 0 ; LBA of 0 s0:0067 push large dword ptr [bp+8] ; DATA XREF: seg000:00E5r s0:0067 ; seg000:0125r s0:0067 ; Transfer buffer s0:0067 ; This is also the LBA of first absolute sector s0:0067 ; in the MBR partition record s0:006B push 0 ; Transfer buffer s0:006E push 7C00h ; Number of blocks s0:006E ; Note that only the first byte is relevant, s0:006E ; and the second is ignored, so we are only s0:006E ; reading 7Ch s0:0071 push 1 ; Reserved s0:0074 push 10h ; Packet is size 10h s0:0077 mov ah, 42h ; 'B' s0:0079 mov dl, [bp+0] ; Drive number s0:007C mov si, sp ; Point to the address packet we just made s0:007E int 13h ; DISK - Extended Read s0:007E ; s0:007E ; Reads DS:SI into a disk appress packet s0:007E ; a disk address packet is: s0:007E ; 00 BYTE: Size of packet (10h or 18h) s0:007E ; 01 BYTE: Reserved s0:007E ; 02 WORD: Number of blocks to transfer s0:007E ; 04 DWORD: Transfer buffer s0:007E ; 08 QWORD: Starting absolute block number (LBA) s0:007E ; 10 QWORD: 64-bit flat address of transfer buffer (optional, used if the DWORD at 04 is FFFFh:FFFFh) s0:007E ; s0:007E ; CF cleared if successful s0:007E ; AH = 0 on success s0:0080 lahf ; Preserve the flags for a second s0:0081 add sp, 10h ; Pop the address packet we were using off the stack s0:0084 sahf ; Store AH into Flags Register s0:0085 jmp short PostDiskReadState ; Jump s0:0087 ; --------------------------------------------------------------------------- s0:0087 s0:0087 InstallationFailed: ; CODE XREF: seg000:005Fj s0:0087 mov ax, 201h ; The extended read interrupt is not installed, s0:0087 ; so use the legacy version s0:0087 ; AH = 2 (Disk read sectors into memory) s0:0087 ; AL = 1 (Read 1 sector) s0:008A mov bx, 7C00h ; This is the destination buffer s0:008D mov dl, [bp+0] ; Drive number s0:0090 mov dh, [bp+1] ; These 3 bytes are a CHS structure s0:0093 mov cl, [bp+2] s0:0096 mov ch, [bp+3] s0:0099 int 13h ; DISK - READ SECTORS INTO MEMORY s0:0099 ; AL = number of sectors to read, CH = track, CL = sector s0:0099 ; DH = head, DL = drive, ES:BX -> buffer to fill s0:0099 ; Return: CF set on error, AH = status, AL = number of sectors read

Right away, a comparison is done against the sentinel value. If it is 0 (indicating no extended read), a jump is taken to the InstallationFailed label. If the extended read is supported, an “address packet” is set up for the interrupt and then the int 13h call is made, performing the read. This was actually a pretty tricky section to figure out, mostly because I found all the documentation about address packets was pretty confusing. After the extended read is completed, the code jumps to the PostDiskReadState location. I expect there are a few errors in my comments about the address packet, which I might try to figure out more about later.

Both the extended read and the non-extended read code do essentially the same thing though. They load the first sector of the bootable partition into memory at 0000h:7C00h. This will most likely be the operating system’s loader which will get things started up properly. Before we jump to the OS though, first we have a little bit of maintenance to do, to make sure we are set up and good to go.

After the disk reads are done, we need to make sure that everything succeeded, which is what this block of code does:

s0:009B PostDiskReadState: ; CODE XREF: seg000:0085j s0:009B popad ; Restore all our registers s0:009D jnb short DiskReadSuccess ; Jump if CF = 0 s0:009D ; i.e. The interrupt we just executed (either one) s0:009D ; just succeeded. s0:009F dec byte ptr [bp+11h] ; Remember this was set to 5 before? s0:009F ; This is looping and trying the disk several times, s0:009F ; (it might have failed while the disk spun up) s0:00A2 jnz short ReattemptDiskRead ; Jump if Not Zero (ZF=0) s0:00A4 cmp byte ptr [bp+0], 80h ; 'Ç' ; Is this drive letter 80h, i.e. the s0:00A4 ; first hard drive? s0:00A8 jz PrintErrorLoadingOperatingSystem ; Jump if Zero (ZF=1) s0:00AC mov dl, 80h ; 'Ç' ; Re-try on the first hard drive s0:00AE jmp short FoundBootableEntry ; Jump s0:00B0 ; --------------------------------------------------------------------------- s0:00B0 s0:00B0 ReattemptDiskRead: ; CODE XREF: seg000:00A2j s0:00B0 push bp s0:00B1 xor ah, ah ; Logical Exclusive OR s0:00B3 mov dl, [bp+0] s0:00B6 int 13h ; DISK - RESET DISK SYSTEM s0:00B6 ; DL = drive (if bit 7 is set both hard disks and s0:00B6 ; floppy disks reset) s0:00B6 ; s0:00B6 ; This is important so we can try the read again s0:00B8 pop bp s0:00B9 jmp short AttemptLoadFromDisk ; Jump

For both style of disk reads, if the operation succeeded, the carry flag will be cleared. As such, there is a JNB instruction that is taken if the load succeeded and jumps to the DiskReadSuccess location. If not, the counter we previously set to 5 is decremented. Remember how I talked about that the disk read might fail sometimes? This code is taking into account the drive being busy, not being spun up, or any other reason by just re-trying a few times. If however, it has failed after the 5 attempts, something is wrong. A comparison is then made to see if we are on the first hard drive by comparing the drive letter we wrote to [BP+0] with 80h. If it is, there is a problem loading the OS, so we print an error message. If not, we change the DL register to 80h, indicating the first hard drive, and try the whole process again.

The ReattemptDiskRead code is pretty straightforward. All it does is resets the disk system and jumps back to the AttemptLoadFromDisk location. Pretty simple, huh?

Hopefully the disk read will succeed though, and the code will get to the DiskReadSuccess location. We are very close to actually jumping to the OS in that case. Here is the code:

s0:00BB DiskReadSuccess: ; CODE XREF: seg000:009Dj s0:00BB cmp word ptr ds:7DFEh, 0AA55h ; We just loaded new code to 7C00h. Check s0:00BB ; to see if it has the bootable signature AA55. s0:00BB ; This signature indicates a VBR, which indicates s0:00BB ; an operating system s0:00C1 jnz short PrintMissingOperatingSystem ; The last two bytes are NOT AA55h s0:00C1 ; so there is no OS. s0:00C1 ; Print an error message. s0:00C3 push word ptr [bp+0] s0:00C6 call CheckKeyboardSystemFlag ; Call Procedure s0:00C9 jnz short CheckForTPM ; Jump if Not Zero (ZF=0) s0:00CB cli ; Disable interrupts for a while s0:00CC mov al, 0D1h ; '-' s0:00CE out 64h, al ; AT Keyboard controller 8042. s0:00CE ; Enables writing the output port s0:00D0 call CheckKeyboardSystemFlag ; Call Procedure s0:00D3 mov al, 0DFh ; '¯' s0:00D5 out 60h, al ; AT Keyboard controller 8042. s0:00D5 ; Enables writing to the status register s0:00D7 call CheckKeyboardSystemFlag ; Call Procedure s0:00DA mov al, 0FFh ; Enable A20 memory line s0:00DC out 64h, al ; AT Keyboard controller 8042. s0:00DC ; Reset the keyboard and start internal diagnostics s0:00DE call CheckKeyboardSystemFlag ; Call Procedure s0:00E1 sti ; Enable interrupts s0:0156 ; ¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦ S U B R O U T I N E ¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦ s0:0156 s0:0156 s0:0156 CheckKeyboardSystemFlag proc near ; CODE XREF: seg000:00C6p s0:0156 ; seg000:00D0p ... s0:0156 sub cx, cx ; CX = 0 s0:0158 s0:0158 CheckByte2: ; CODE XREF: CheckKeyboardSystemFlag+8j s0:0158 in al, 64h ; AT Keyboard controller 8042. s0:015A jmp short $+2 ; Jump s0:015C and al, 2 ; Logical AND s0:015E loopne CheckByte2 ; Loop while rCX != 0 and ZF=0 s0:0160 and al, 2 ; Logical AND s0:0162 retn ; Return Near from Procedure s0:0162 CheckKeyboardSystemFlag endp

This code first checks to see that we did, in fact load an OS and not just some garbage into memory by checking for the AA55h signature. If it is not there, we print an error message. Otherwise, we’re going to fiddle with the keyboard controller a bit. This code makes several calls to the CheckKeyboardSystemFlag function, which basically loops until the keyboard controller is ready to talk to to the CPU. I’m a little fuzzy on what the output to the ports is doing, but I’m pretty sure that it is enabling the A20 address line, which enables larger amounts of memory to be used. It’s a pretty common task and there are write-ups all over the Internet, so I won’t go into it.

After the MBR is done fiddling with the keyboard controller, it decides to check for a Trusted Platform Module. The TPM is used to do several security related things, such as for Windows BitLocker encryption. There is a lot of complex documentation, so after I figured out that this was TPM code, I decided not to investigate it further. It is listed below:

s0:00E2 CheckForTPM: ; CODE XREF: seg000:00C9j s0:00E2 mov ax, 0BB00h s0:00E5 int 1Ah s0:00E7 and eax, eax ; Logical AND s0:00EA jnz short JumpToLoadedMemory ; Jump if Not Zero (ZF=0) s0:00EC cmp ebx, 41504354h ; Compare Two Operands s0:00F3 jnz short JumpToLoadedMemory ; Jump if Not Zero (ZF=0) s0:00F5 cmp cx, 102h ; Compare Two Operands s0:00F9 jb short JumpToLoadedMemory ; Jump if Below (CF=1) s0:00FB push large 0BB07h s0:0101 push large 200h s0:0107 push large 8 s0:010D push ebx s0:010F push ebx s0:0111 push ebp s0:0113 push large 0 s0:0119 push large 7C00h s0:011F popad ; Pop all General Registers (use32) s0:0121 push 0 s0:0124 pop es s0:0125 int 1Ah ; This makes a call to the TPM

The code first checks for the presence of a TPM module. If it is not there, it jumps to the JumpToLoadedMemory location, otherwise it makes a call to the TPM.

Well, ladies and gentlement, thanks for sticking with me. After all that, we are finally ready to jump to the operating system that we loaded into memory. It’s very simple, so without further adieu:

s0:0127 JumpToLoadedMemory: ; CODE XREF: seg000:00EAj s0:0127 ; seg000:00F3j ... s0:0127 pop dx s0:0128 xor dh, dh ; Logical Exclusive OR s0:012A jmp far ptr 0:7C00h ; Jump to the code we have loaded

That’s it?! Yup, that’s it. Anti-climactic, though I guess Master Boot Records aren’t supposed to be entertaining.

There is some more code that I didn’t talk about yet, because it has to do with printing the error messages out. I’m not going to explain it here, since I’m already over 3000 words and I’m sure you’re sick of reading. Here it is:

s0:0131 PrintMissingOperatingSystem: ; CODE XREF: seg000:00C1j s0:0131 mov al, ds:7B7h s0:0134 jmp short DisplayErrorMessage ; Jump s0:0136 ; --------------------------------------------------------------------------- s0:0136 s0:0136 PrintErrorLoadingOperatingSystem: ; CODE XREF: seg000:00A8j s0:0136 mov al, ds:7B6h s0:0139 jmp short DisplayErrorMessage ; Jump s0:013B ; --------------------------------------------------------------------------- s0:013B s0:013B PrintInvalidPartitionTable: ; CODE XREF: seg000:0029j s0:013B mov al, ds:7B5h s0:013E s0:013E DisplayErrorMessage: ; CODE XREF: seg000:0134j s0:013E ; seg000:0139j s0:013E xor ah, ah ; Clear out the high byte of the AX register. s0:0140 add ax, 700h ; We now point to 700h + whatever offset we were given s0:0140 ; above. s0:0143 mov si, ax s0:0145 s0:0145 PrintErrorStringLoop: ; CODE XREF: seg000:0151j s0:0145 lodsb ; Load byte at DS:SI into AL s0:0146 cmp al, 0 ; Is the next byte 0? We're looking at a 0 terminated s0:0146 ; string, so this is important. s0:0148 jz short HaltSystem ; Jump if Zero (ZF=1) s0:014A mov bx, 7 s0:014D mov ah, 0Eh s0:014F int 10h ; - VIDEO - WRITE CHARACTER AND ADVANCE CURSOR (TTY WRITE) s0:014F ; AL = character, BH = display page (alpha modes) s0:014F ; BL = foreground color (graphics modes) s0:0151 jmp short PrintErrorStringLoop ; Jump s0:0153 ; --------------------------------------------------------------------------- s0:0153 s0:0153 HaltSystem: ; CODE XREF: seg000:0148j s0:0153 ; seg000:0154j s0:0153 hlt ; This stops the computer. s0:0154 jmp short HaltSystem ; Jump s0:0156 s0:0162 ; --------------------------------------------------------------------------- s0:0163 aInvalidPartiti db 'Invalid partition table',0 s0:017B aErrorLoadingOp db 'Error loading operating system',0 s0:019A aMissingOperati db 'Missing operating system',0 s0:01B3 db 0 s0:01B4 db 0 s0:01B5 db 63h ; c ; Redirect to the error message at 63h, s0:01B5 ; i.e. "Invalid Partition Table" s0:01B6 db 7Bh ; { ; Redirect to the error message at 7Bh, s0:01B6 ; i.e. "Error loading operating system" s0:01B7 db 9Ah ; Ü ; Redirect to the error message at 9Ah, s0:01B7 ; i.e. "Missing operating system"

I realize that a blog post is a pretty difficult way to explain an RE task like this well. As such, I’m going to make my IDA Pro database available for download here. If you don’t have IDA, there is a free version (which I use) available here. It is missing quite a few features, but the price is right, and for work like this (16-bit MBR code), a lot of the newer features aren’t even relevant.

If you’re interested, you could do this analysis on your own computer. Depending on your computer’s manufacturer, you might have some small differences, but those are what makes it fun, right? To get started, get the trial of Hex Workshop, extract the MBR (it’s the first sector on the disk), then use the free version of IDA Pro to examine it. Happy hunting!