Is there any special reason for STOSB to use extra segment? [closed]
I have read that STOSB functions like this:
ES:[DI] <-- AL
If DF = 0 increment DI else decrement DI.
- So why
STOSBdoesn't changeDS:[DI]? - Is there a special purpose for using extra segment?
- In most string instructions we use extra segment. Why?
assembly x86-16
edited Dec 30 '18 at 15:56
Sep Roland
11.9k21945
asked Dec 29 '18 at 14:46
oscar tabarezoscar tabarez
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closed as too broad by old_timer, Roman Pokrovskij, Cindy Meister, Umair, Kaushik Nayak Dec 31 '18 at 8:12
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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I have read that STOSB functions like this:
ES:[DI] <-- AL
If DF = 0 increment DI else decrement DI.
- So why
STOSBdoesn't changeDS:[DI]? - Is there a special purpose for using extra segment?
- In most string instructions we use extra segment. Why?
assembly x86-16
edited Dec 30 '18 at 15:56
Sep Roland
11.9k21945
asked Dec 29 '18 at 14:46
oscar tabarezoscar tabarez
475
closed as too broad by old_timer, Roman Pokrovskij, Cindy Meister, Umair, Kaushik Nayak Dec 31 '18 at 8:12
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I have read that STOSB functions like this:
ES:[DI] <-- AL
If DF = 0 increment DI else decrement DI.
- So why
STOSBdoesn't changeDS:[DI]? - Is there a special purpose for using extra segment?
- In most string instructions we use extra segment. Why?
assembly x86-16
edited Dec 30 '18 at 15:56
Sep Roland
11.9k21945
asked Dec 29 '18 at 14:46
oscar tabarezoscar tabarez
475
I have read that STOSB functions like this:
ES:[DI] <-- AL
If DF = 0 increment DI else decrement DI.
- So why
STOSBdoesn't changeDS:[DI]? - Is there a special purpose for using extra segment?
- In most string instructions we use extra segment. Why?
assembly x86-16
assembly x86-16
edited Dec 30 '18 at 15:56
Sep Roland
11.9k21945
asked Dec 29 '18 at 14:46
oscar tabarezoscar tabarez
475
edited Dec 30 '18 at 15:56
Sep Roland
11.9k21945
asked Dec 29 '18 at 14:46
oscar tabarezoscar tabarez
475
edited Dec 30 '18 at 15:56
Sep Roland
11.9k21945
edited Dec 30 '18 at 15:56
Sep Roland
11.9k21945
edited Dec 30 '18 at 15:56
Sep Roland
11.9k21945
11.9k21945
asked Dec 29 '18 at 14:46
oscar tabarezoscar tabarez
475
asked Dec 29 '18 at 14:46
oscar tabarezoscar tabarez
475
asked Dec 29 '18 at 14:46
oscar tabarezoscar tabarez
475
475
closed as too broad by old_timer, Roman Pokrovskij, Cindy Meister, Umair, Kaushik Nayak Dec 31 '18 at 8:12
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as too broad by old_timer, Roman Pokrovskij, Cindy Meister, Umair, Kaushik Nayak Dec 31 '18 at 8:12
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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So why STOSB doesn't change DS:[DI]
Because this definition would collide with the use of LODSB which already uses DS:[SI]. Using a separate segment register gives you more flexibility.
Is there a special purpose for using extra segment?
Yes. You can transfer bytes between segments easily while processing them. For example, you can use a LODSB loading AL from DS:[SI], modify AL and then store it to a different segment, the Extra Segment, with STOSB using ES:[DI]. In 8086 with its 16-bit segment size and a 20-bit address space this is really useful.
Another instruction illustrating the use is the REP MOVSB instruction which copies a sequence of bytes (with its length in CX) from DS:[SI] to ES:[DI].
(If you don't need to examine each byte as you copy it, you'd simply use rep movsb or rep movsw for better performance than lods/stos in a loop.)
In most string instructions we use extra segment. why?
Well, not in most, but maybe roughly in half of them. Using another segment register gives you the advantage of quick access to different segments - you are not limited to processing data only in one 64KB segment and do not have to change the DS register before each access to a different segment.
stos and movs write es:[di], which makes sense because DI is the "destination index" register.
cmps and scas read from es:[di], which is maybe surprising for scas because it only has one memory operand so you might expect it to read from ds:[si] like lods. Especially because SCASB sets flags from AL - [mem], not the other way around, so it's like a cmp where memory is the right operand (source), not left (destination). Like cmp al, es:[di].
Perhaps the architect of the 8086's instruction set imagined a use-case of a loop that does lods and scas to implement strcmp between segments.
edited Dec 30 '18 at 15:49
Sep Roland
11.9k21945
answered Dec 29 '18 at 15:06
zx485zx485
14k122947
For most of the string instructions, ES:DI (DI -> destination index) is the store address, with SCASB/W being the exception that is uses ES:DI to scan (read) memory for a matching byte / word (in 16 bit mode).
– rcgldr
Dec 29 '18 at 15:29
If your length is ECX (not just CX), your addresses will also be 32-bit DS:[ESI], not SI. There might be a combination of prefixes which could achieve 16-bit address-size and 32-bit counter size in some mode, but I'm not sure. It certainly wouldn't be normal, though.
– Peter Cordes
Dec 30 '18 at 8:49
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
So why STOSB doesn't change DS:[DI]
Because this definition would collide with the use of LODSB which already uses DS:[SI]. Using a separate segment register gives you more flexibility.
Is there a special purpose for using extra segment?
Yes. You can transfer bytes between segments easily while processing them. For example, you can use a LODSB loading AL from DS:[SI], modify AL and then store it to a different segment, the Extra Segment, with STOSB using ES:[DI]. In 8086 with its 16-bit segment size and a 20-bit address space this is really useful.
Another instruction illustrating the use is the REP MOVSB instruction which copies a sequence of bytes (with its length in CX) from DS:[SI] to ES:[DI].
(If you don't need to examine each byte as you copy it, you'd simply use rep movsb or rep movsw for better performance than lods/stos in a loop.)
In most string instructions we use extra segment. why?
Well, not in most, but maybe roughly in half of them. Using another segment register gives you the advantage of quick access to different segments - you are not limited to processing data only in one 64KB segment and do not have to change the DS register before each access to a different segment.
stos and movs write es:[di], which makes sense because DI is the "destination index" register.
cmps and scas read from es:[di], which is maybe surprising for scas because it only has one memory operand so you might expect it to read from ds:[si] like lods. Especially because SCASB sets flags from AL - [mem], not the other way around, so it's like a cmp where memory is the right operand (source), not left (destination). Like cmp al, es:[di].
Perhaps the architect of the 8086's instruction set imagined a use-case of a loop that does lods and scas to implement strcmp between segments.
edited Dec 30 '18 at 15:49
Sep Roland
11.9k21945
answered Dec 29 '18 at 15:06
zx485zx485
14k122947
For most of the string instructions, ES:DI (DI -> destination index) is the store address, with SCASB/W being the exception that is uses ES:DI to scan (read) memory for a matching byte / word (in 16 bit mode).
– rcgldr
Dec 29 '18 at 15:29
If your length is ECX (not just CX), your addresses will also be 32-bit DS:[ESI], not SI. There might be a combination of prefixes which could achieve 16-bit address-size and 32-bit counter size in some mode, but I'm not sure. It certainly wouldn't be normal, though.
– Peter Cordes
Dec 30 '18 at 8:49
add a comment |
So why STOSB doesn't change DS:[DI]
Because this definition would collide with the use of LODSB which already uses DS:[SI]. Using a separate segment register gives you more flexibility.
Is there a special purpose for using extra segment?
Yes. You can transfer bytes between segments easily while processing them. For example, you can use a LODSB loading AL from DS:[SI], modify AL and then store it to a different segment, the Extra Segment, with STOSB using ES:[DI]. In 8086 with its 16-bit segment size and a 20-bit address space this is really useful.
Another instruction illustrating the use is the REP MOVSB instruction which copies a sequence of bytes (with its length in CX) from DS:[SI] to ES:[DI].
(If you don't need to examine each byte as you copy it, you'd simply use rep movsb or rep movsw for better performance than lods/stos in a loop.)
In most string instructions we use extra segment. why?
Well, not in most, but maybe roughly in half of them. Using another segment register gives you the advantage of quick access to different segments - you are not limited to processing data only in one 64KB segment and do not have to change the DS register before each access to a different segment.
stos and movs write es:[di], which makes sense because DI is the "destination index" register.
cmps and scas read from es:[di], which is maybe surprising for scas because it only has one memory operand so you might expect it to read from ds:[si] like lods. Especially because SCASB sets flags from AL - [mem], not the other way around, so it's like a cmp where memory is the right operand (source), not left (destination). Like cmp al, es:[di].
Perhaps the architect of the 8086's instruction set imagined a use-case of a loop that does lods and scas to implement strcmp between segments.
edited Dec 30 '18 at 15:49
Sep Roland
11.9k21945
answered Dec 29 '18 at 15:06
zx485zx485
14k122947
For most of the string instructions, ES:DI (DI -> destination index) is the store address, with SCASB/W being the exception that is uses ES:DI to scan (read) memory for a matching byte / word (in 16 bit mode).
– rcgldr
Dec 29 '18 at 15:29
If your length is ECX (not just CX), your addresses will also be 32-bit DS:[ESI], not SI. There might be a combination of prefixes which could achieve 16-bit address-size and 32-bit counter size in some mode, but I'm not sure. It certainly wouldn't be normal, though.
– Peter Cordes
Dec 30 '18 at 8:49
add a comment |
So why STOSB doesn't change DS:[DI]
Because this definition would collide with the use of LODSB which already uses DS:[SI]. Using a separate segment register gives you more flexibility.
Is there a special purpose for using extra segment?
Yes. You can transfer bytes between segments easily while processing them. For example, you can use a LODSB loading AL from DS:[SI], modify AL and then store it to a different segment, the Extra Segment, with STOSB using ES:[DI]. In 8086 with its 16-bit segment size and a 20-bit address space this is really useful.
Another instruction illustrating the use is the REP MOVSB instruction which copies a sequence of bytes (with its length in CX) from DS:[SI] to ES:[DI].
(If you don't need to examine each byte as you copy it, you'd simply use rep movsb or rep movsw for better performance than lods/stos in a loop.)
In most string instructions we use extra segment. why?
Well, not in most, but maybe roughly in half of them. Using another segment register gives you the advantage of quick access to different segments - you are not limited to processing data only in one 64KB segment and do not have to change the DS register before each access to a different segment.
stos and movs write es:[di], which makes sense because DI is the "destination index" register.
cmps and scas read from es:[di], which is maybe surprising for scas because it only has one memory operand so you might expect it to read from ds:[si] like lods. Especially because SCASB sets flags from AL - [mem], not the other way around, so it's like a cmp where memory is the right operand (source), not left (destination). Like cmp al, es:[di].
Perhaps the architect of the 8086's instruction set imagined a use-case of a loop that does lods and scas to implement strcmp between segments.
edited Dec 30 '18 at 15:49
Sep Roland
11.9k21945
answered Dec 29 '18 at 15:06
zx485zx485
14k122947
So why STOSB doesn't change DS:[DI]
Because this definition would collide with the use of LODSB which already uses DS:[SI]. Using a separate segment register gives you more flexibility.
Is there a special purpose for using extra segment?
Yes. You can transfer bytes between segments easily while processing them. For example, you can use a LODSB loading AL from DS:[SI], modify AL and then store it to a different segment, the Extra Segment, with STOSB using ES:[DI]. In 8086 with its 16-bit segment size and a 20-bit address space this is really useful.
Another instruction illustrating the use is the REP MOVSB instruction which copies a sequence of bytes (with its length in CX) from DS:[SI] to ES:[DI].
(If you don't need to examine each byte as you copy it, you'd simply use rep movsb or rep movsw for better performance than lods/stos in a loop.)
In most string instructions we use extra segment. why?
Well, not in most, but maybe roughly in half of them. Using another segment register gives you the advantage of quick access to different segments - you are not limited to processing data only in one 64KB segment and do not have to change the DS register before each access to a different segment.
stos and movs write es:[di], which makes sense because DI is the "destination index" register.
cmps and scas read from es:[di], which is maybe surprising for scas because it only has one memory operand so you might expect it to read from ds:[si] like lods. Especially because SCASB sets flags from AL - [mem], not the other way around, so it's like a cmp where memory is the right operand (source), not left (destination). Like cmp al, es:[di].
Perhaps the architect of the 8086's instruction set imagined a use-case of a loop that does lods and scas to implement strcmp between segments.
edited Dec 30 '18 at 15:49
Sep Roland
11.9k21945
answered Dec 29 '18 at 15:06
zx485zx485
14k122947
edited Dec 30 '18 at 15:49
Sep Roland
11.9k21945
edited Dec 30 '18 at 15:49
Sep Roland
11.9k21945
edited Dec 30 '18 at 15:49
Sep Roland
11.9k21945
11.9k21945
answered Dec 29 '18 at 15:06
zx485zx485
14k122947
answered Dec 29 '18 at 15:06
zx485zx485
14k122947
answered Dec 29 '18 at 15:06
zx485zx485
14k122947
14k122947
For most of the string instructions, ES:DI (DI -> destination index) is the store address, with SCASB/W being the exception that is uses ES:DI to scan (read) memory for a matching byte / word (in 16 bit mode).
– rcgldr
Dec 29 '18 at 15:29
If your length is ECX (not just CX), your addresses will also be 32-bit DS:[ESI], not SI. There might be a combination of prefixes which could achieve 16-bit address-size and 32-bit counter size in some mode, but I'm not sure. It certainly wouldn't be normal, though.
– Peter Cordes
Dec 30 '18 at 8:49
add a comment |
For most of the string instructions, ES:DI (DI -> destination index) is the store address, with SCASB/W being the exception that is uses ES:DI to scan (read) memory for a matching byte / word (in 16 bit mode).
– rcgldr
Dec 29 '18 at 15:29
If your length is ECX (not just CX), your addresses will also be 32-bit DS:[ESI], not SI. There might be a combination of prefixes which could achieve 16-bit address-size and 32-bit counter size in some mode, but I'm not sure. It certainly wouldn't be normal, though.
– Peter Cordes
Dec 30 '18 at 8:49
For most of the string instructions, ES:DI (DI -> destination index) is the store address, with SCASB/W being the exception that is uses ES:DI to scan (read) memory for a matching byte / word (in 16 bit mode).
– rcgldr
Dec 29 '18 at 15:29
For most of the string instructions, ES:DI (DI -> destination index) is the store address, with SCASB/W being the exception that is uses ES:DI to scan (read) memory for a matching byte / word (in 16 bit mode).
– rcgldr
Dec 29 '18 at 15:29
If your length is ECX (not just CX), your addresses will also be 32-bit DS:[ESI], not SI. There might be a combination of prefixes which could achieve 16-bit address-size and 32-bit counter size in some mode, but I'm not sure. It certainly wouldn't be normal, though.
– Peter Cordes
Dec 30 '18 at 8:49
If your length is ECX (not just CX), your addresses will also be 32-bit DS:[ESI], not SI. There might be a combination of prefixes which could achieve 16-bit address-size and 32-bit counter size in some mode, but I'm not sure. It certainly wouldn't be normal, though.
– Peter Cordes
Dec 30 '18 at 8:49
add a comment |
