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Armature wrote:

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Boot:
	cli
	mov ax, 0x0000
	mov es, ax
	mov dx, ax
	mov ss, ax
	mov sp, 0x7C00
	mov bp, 0x0500
	sti

Armature wrote:

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ReadDisk:
	mov ah, 0x02
	mov al, 13	;how many sectors to read
	mov cl, 19	;Starting sector
	mov ch, 0	;cylinder
	mov dh, 0	;head
	mov dl, 0x80	;Floppy Drive

Armature wrote:

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	mov bx, 0x7E00	;ES:BX 0x0000:0x7E00
	int 0x13
	jc _DiskReadError
	mov cx, 8
	mov di, 0x7E00
	mov si, KERNEL
	jmp ReadFAT

_DiskReadError:
	mov si, READERR
	jmp BPrint

ReadFAT:
	cmp si, di
	jne _NotFound
	inc si
	inc di
	dec cx
	cmp cx, 0
	je _LoadEntry
	jne ReadFAT

Armature wrote:I then write the code to the boot sector of a floppy drive and insert the floppy drive into a windows computer but the computer reports back that the disk doesn't have a filesystem.

Octocontrabass wrote:

Armature wrote:I then write the code to the boot sector of a floppy drive and insert the floppy drive into a windows computer but the computer reports back that the disk doesn't have a filesystem.

Use a hex editor to compare your boot sector with a working boot sector. You've probably made a mistake somewhere in the BPB.

BenLunt wrote:

Armature wrote:

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Boot:
	cli
	mov ax, 0x0000
	mov es, ax
	mov dx, ax
	mov ss, ax
	mov sp, 0x7C00
	mov bp, 0x0500
	sti

I am guessing that the "dx" above should have been "ds".

Armature wrote:

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ReadDisk:
	mov ah, 0x02
	mov al, 13	;how many sectors to read
	mov cl, 19	;Starting sector
	mov ch, 0	;cylinder
	mov dh, 0	;head
	mov dl, 0x80	;Floppy Drive

A value of 0x80 is not a floppy drive... (hint: dl < 0x80 == floppy. dl >= 0x80 == hard drive)

Armature wrote:

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	mov bx, 0x7E00	;ES:BX 0x0000:0x7E00
	int 0x13
	jc _DiskReadError
	mov cx, 8
	mov di, 0x7E00
	mov si, KERNEL
	jmp ReadFAT

_DiskReadError:
	mov si, READERR
	jmp BPrint

ReadFAT:
	cmp si, di
	jne _NotFound
	inc si
	inc di
	dec cx
	cmp cx, 0
	je _LoadEntry
	jne ReadFAT

Does the loop above actually do anything? si and di will never be the same. If it was, it will loop until cx == 0 and then go to _LoadEntry. However, since it doesn't, it goes directly to _NotFound anyway. Plus, you are checking that the address of KERNEL and 0x7E00 ever match. You need to be checking the two bytes *at* those addresses.

Your code actually does nothing after it "tries" to read 13 sectors from the *end of* the 1st track. Head = 0, Cyl = 0, Sector = 19 is an non-existant sector on a 1.44 standard floppy disk. This type of floppy has 80 cylinders, 2 sides per cylinder (tracks), 18 sectors per track = 2880 sectors.

I have an out-dated version of my FAT boot sector at https://github.com/fysnet/FYSOS/blob/ma ... at1x_f.asm that you might wish to study. I warn you though, if you simply copy and paste it, you will not get what you think you are getting. i.e.: it won't work right out of the box. You must study it and learn what it is doing so that you can adapt it, or better yet, write your own to work.

Ben
- http://www.fysnet.net/osdesign_book_series.htm

Armature wrote:Ive figured out what the issue was. I was jumping four bytes using

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jmp near boot
nop

instead of the three using

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jmp short boot
nop

BenLunt wrote:

Armature wrote:Ive figured out what the issue was. I was jumping four bytes using

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jmp near boot
nop

instead of the three using

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jmp short boot
nop

Good. Something specific to the assembler, but still, good. However, does it actually boot? I mean, does it find the next stage (kernel in your case), load, and jump to it?

Ben

Armature wrote:I just tried to find the kernel on the floppy rive using the code but it didn't find it. I did change the code to this :

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org 0x7C00
jmp near Boot
nop

Armature wrote:

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BPrint:
	mov ah, 0x0E
	_BPrintLoop:
	lodsb
	cmp al, 0
	je _BDone
	int 0x10
	jmp _BPrintLoop
	_BDone:
	ret

Armature wrote:

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ReadDisk:
	mov ah, 0x02
	mov al, 13	;how many sectors to read
	mov cl, 1	;Starting sector
	mov ch, 1	;cylinder
	mov dh, 0	;head
	mov dl, 0x00	
	mov bx, 0x7E00	;ES:BX 0x0000:0x7E00
	int 0x13

Armature wrote:

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ReadFAT:
	mov di, [0x7E00]
	mov si, [KERNEL]
	cmp si, di
	jne _NotFound
	inc si
	inc di

Armature wrote:Hey, I've read through your code and some other tutorials on how to convert LBA to CHS but I'm left really confused. I understand the maths behind the code and how the CHS are calculated but what I don't understand is the storing of the variables calculated. The outputs are stored in 16 bit registers when the interrupt using the 8bit registers to read and write data. Would you be able to explain this to me please?

 CH = low eight bits of cylinder number
 CL = sector number 1-63 (bits 0-5)
         high two bits of cylinder (bits 6-7, hard disk only)
 DH = head number

  11111100 00000000
  54321098 76543210
  CCCCCCCC CCSSSSSS

  76543210 98xxxxxx
  CCCCCCCC CCxxxxxx

  ; CX = 10-bit cylinder number in CHS form (zero based)
  mov  dh,cl
  shr  dh,6
  mov dl,ch
 ; DX now is 10-bit cylinder number (zero based)

  ; DX = 10-bit cylinder number (zero based)
  ; AL = 6-bit sector number (1 based)
  mov  ch,dl
  mov cl,dh
  shl  cl,6
  and  al,3Fh
  or  cl,al
  ; CX now is 10-bit cylinder number in CHS form (zero based)

BenLunt wrote:

Armature wrote:Hey, I've read through your code and some other tutorials on how to convert LBA to CHS but I'm left really confused. I understand the maths behind the code and how the CHS are calculated but what I don't understand is the storing of the variables calculated. The outputs are stored in 16 bit registers when the interrupt using the 8bit registers to read and write data. Would you be able to explain this to me please?

For Floppy Drives, 8-bit registers only might be okay. However, CHS (Cylinder/Head/Sector) values are stored in three 8-bit registers, though each 8-bit register is *not* exclusive to only one of the values.

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 CH = low eight bits of cylinder number
 CL = sector number 1-63 (bits 0-5)
         high two bits of cylinder (bits 6-7, hard disk only)
 DH = head number

The indication that the high two bits are hard-drives only is simply because a floppy disk doesn't have that many cylinders. You still have to place the high two bits of the cylinder (which for floppy disks will be 00b) into the high two bits of CL.

Let's show an example. CX is the 16-bit register, consisting of CH and CL. Therefore, shown is the 16-bit representation of the cylinder and sector bits: (assuming that your version of this forum uses fixed width fonts for all code blocks)

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  11111100 00000000
  54321098 76543210
  CCCCCCCC CCSSSSSS

Where 'C' = Cylinder and 'S' = Sector. However, the 10-bit cylinder value is actually:

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  76543210 98xxxxxx
  CCCCCCCC CCxxxxxx

To get the actual Cylinder number from CX above, you can do something like:

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  ; CX = 10-bit cylinder number in CHS form (zero based)
  mov  dh,cl
  shr  dh,6
  mov dl,ch
 ; DX now is 10-bit cylinder number (zero based)

and DX will now contain a valid 10-bit cylinder number. Convert from a 10-bit number back to CHS, you could do something like:

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  ; DX = 10-bit cylinder number (zero based)
  ; AL = 6-bit sector number (1 based)
  mov  ch,dl
  mov cl,dh
  shl  cl,6
  and  al,3Fh
  or  cl,al
  ; CX now is 10-bit cylinder number in CHS form (zero based)

The code that I directed you to might not be near as clear on what it is doing, but this is because a boot sector needs all of the room it can get. For boot sectors, if you can do the conversion in 16 bytes and have unclear code, compared to 32 bytes to have clear code, you do the 16-byte version and include a well-documented comment block indicating what you are doing.

Does this help?
Ben