Index
> DOS > Segments |
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typedef 31 Dec 2011, 05:01
Ds:si is address and offset all put together.
It's like accessing an element in an array. In this case DS is the element's base address. SI is the index/location/offset. Here's a good example: DS = book shelf SI = cabinet The same for the latter. Hope that helps |
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31 Dec 2011, 05:01 |
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GhostXoPCorp 31 Dec 2011, 05:47
Thank you. The problem was i was using movsb, when i should of been using lodsb.
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31 Dec 2011, 05:47 |
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typedef 31 Dec 2011, 20:21
They both work off the same registers.
The latter one moves a byte from SI to DI. You must set the length of the bytes in CX before calling it, if you want it to handle the copying for you. CX specifies how many times to copy bytes from the source address. |
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31 Dec 2011, 20:21 |
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GhostXoPCorp 31 Dec 2011, 21:38
I dont know how, but i added code to try and correct the segments to get movsb to work. The second i took it off the code.. it worked.
Very useful, its been forever since ive done assember (took a break and did some C++ in game development) it was actually fun, but ive found assembler to be alittle more fun, and more controlling  _________________ Oh that divide overflow. Just jumps out of the bushes every time to scare the day lights out of me. |
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31 Dec 2011, 21:38 |
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GhostXoPCorp 31 Dec 2011, 22:15
Would there ever be a time when i have to set the ds and es segments myself (not being the operating system) but in a regular program
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31 Dec 2011, 22:15 |
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GhostXoPCorp 31 Dec 2011, 22:19
i would imagine, the operating system sets the segments, and all i have to do is do lea si, var (if its not a com program) to get the offset of var in my data segment. But in flat binary (one segment) i dont have to worry about segments?
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31 Dec 2011, 22:19 |
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Coty 31 Dec 2011, 22:39
stosb moves a byte from AX to es:di
lodsb moves a byte to AX from ds:si Quote: Would there ever be a time when i have to set the ds and es segments myself (not being the operating system) but in a regular program Yes, let's say DOS loads you at 0x4000:0x0100, video memory is located at 0xA000:0x0000 (for video mode) if you want to directly draw to the display buffer (It's the fastest way) then this is how you would do it: Code: ; Paint a white pixle on the screen... use16 org 0x0100 mov al, 0x13 ; Set video mode '13h' int 0x10 push 0xA000 ; ES == 0xA000 pop es mov di, 0 ; DI == 0x0000 mov al, 0xFF ; AL color = white... stosb ; Right byte in al jmp $ If you don't understand WHY x86 uses segment:offset, it works like this. A 16 bit CPU can only access 64kb of RAM, so, the x86 had an extended buss, that you access the memory in 64kb chunks. Code: {[64kb chunk][64kb chunk][64kb chunk][64kb chunk][64kb chunk][64kb chunk]} es/ds lets you choose what 64kb chunk to read/right from and si/di let you choose the byte in the 64kb chunk... Also on a side note the processor has CS:IP, this is how the cpu knows that memorychunk to exicute, and IP holds what instruction to exicute in that 64kb chink, but you need not worry about this... Last edited by Coty on 01 Jan 2012, 00:01; edited 1 time in total |
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31 Dec 2011, 22:39 |
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typedef 31 Dec 2011, 23:04
$FFFF > sizeof(AL)
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31 Dec 2011, 23:04 |
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GhostXoPCorp 31 Dec 2011, 23:28
Thank you. A fully explained answer, with an example.
Im happy to come back and see you've gotten pretty far in this coty, anyone who works this hard to learn deserves recognition. Good job, keep it up. _________________ Oh that divide overflow. Just jumps out of the bushes every time to scare the day lights out of me. |
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31 Dec 2011, 23:28 |
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Coty 01 Jan 2012, 00:00
Ahahaha! typedef is correct! I have fixed the example...
@GhostXoPCorp: Thanks, best of luck to you! |
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01 Jan 2012, 00:00 |
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GhostXoPCorp 01 Jan 2012, 00:08
Yeah 65535 > 255
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01 Jan 2012, 00:08 |
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AsmGuru62 01 Jan 2012, 11:54
@GhostXoPCorp: if you use flat memory model, like:
- MS-DOS COM file - Win32 PE file Then you do not need to set ES,DS -- all segments are set by the OS in these cases. One small exception, however, if you allocate memory in MS-DOS (COM files can do it by resizing their own image in memory and then using INT 21H function 48H to allocate dynamically) -- in that case you DO need to set ES or DS to the segment returned by function 48H. After using the allocated memory - you need to restore the DS back to the original value given by DOS, otherwise, the global variables declared in COM file will not be accessible. |
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01 Jan 2012, 11:54 |
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GhostXoPCorp 01 Jan 2012, 19:22
I know, ive also taken a look at the MZ Exe format as well, this basically gave me a heads up for things to come. Thank you all for helping me understand this in all reality its the only thing i dont understand, but if i have anymore questions ill be back.
Happy new year  _________________ Oh that divide overflow. Just jumps out of the bushes every time to scare the day lights out of me. |
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01 Jan 2012, 19:22 |
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freecrac 02 Jan 2012, 08:35
 A little mistake has slept in.
Coty wrote: stosb moves a byte from AX to es:di stosb moves a byte from AL to es:di Quote: lodsb moves a byte to AX from ds:si lodsb moves a byte to AL from ds:si Additional the Offsetaddress in DI and/or SI will be increase/decrease by 1 in correlation of the direction flag. ... If we want to use AX = 16 bit (AH + AL): stosw lodsw scasw cmpsw movsw (Offset(s) will be will be increase/decrease by 2) ... On 80386+ we can use EAX = 32 bit: stosd lodsd scasd cmpsd movsd (Offset(s) will be will be increase/decrease by 4.) ... In combination with the repeat-prefix the string instruction repeats the times we have specified in the count register ((E)CX). (Offsets will be increase/decrease in correlation of the direction flag.) rep stosb/w/d rep lodsb/w/d rep scasb/w/d rep cmpsb/w/d rep movsb/w/d Happy new year 
Dirk |
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02 Jan 2012, 08:35 |
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GhostXoPCorp 04 Jan 2012, 04:03
Thanks, but as assumed, default in my opinion is
if a byte is moved from a 16 bit (4 nibble, 2 byte blah blah) register By default IMO means it was moved from al (the lower 8 bits, lower 2 nibbles, lower byte) I know basically all i need to about assembler, just a bit about segments i didnt know, that i now fully understand. Thank you for your help however and happy new year to you as well! _________________ Oh that divide overflow. Just jumps out of the bushes every time to scare the day lights out of me. |
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04 Jan 2012, 04:03 |
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freecrac 04 Jan 2012, 09:43
GhostXoPCorp wrote: I know basically all i need to about assembler, just a bit about segments i didnt know, that i now fully understand. On 80386+ we can enlarge the segmentsize frrom 64 KiB up to 4 GiB. This can be done in our GDT/LDT. Now we can access any address of 4 GiB using a pair of one Segmentregister and a 32 Bit Offsetaddress. AsmGuru62 wrote: ...if you use flat memory model, like: But if we want to set the address of the segment higher than 0 using a linear access up to 4 Gib, then we have to recalculate and decrease our Offsetaddress, so that the address (maybe to the linear frame buffer) is well pointed. Exambple: Code: xor eax, eax mov ax, ds mov edi, LFB_ADDRESS shl eax, 4 sub edi, eax xor esi, esi ; pointed to the begin of our datasegment mov eax, [esi] ; get a pixelcolor from our datasegment using DS:ESI mov [edi], eax ; set the pixel in the upper left corner of the screen using DS:EDI Dirk |
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04 Jan 2012, 09:43 |
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