What is the meaning of M in the Z80 statement ADD A,M
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Recently I have been trying to compile CP/M 2.2 from source. When I try to assemble, everything works except for the instructions ADD A,M and SBC A,M, which the assembler returns a syntax error. I have not done much Z80, so I am a little confused on what the source code means by M?
I am using the CP/M 2.2 Source (Z80 Mnemonics) from the Unofficial CP/M Website.
z80 assembly cp-m
edited 3 hours ago
Alex Hajnal
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asked 5 hours ago
tergav17
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Recently I have been trying to compile CP/M 2.2 from source. When I try to assemble, everything works except for the instructions ADD A,M and SBC A,M, which the assembler returns a syntax error. I have not done much Z80, so I am a little confused on what the source code means by M?
I am using the CP/M 2.2 Source (Z80 Mnemonics) from the Unofficial CP/M Website.
z80 assembly cp-m
edited 3 hours ago
Alex Hajnal
3,45031632
asked 5 hours ago
tergav17
61
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tergav17 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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It might be helpful if you link the source in question and explain where you acquired it.
– Raffzahn
5 hours ago
add a comment |
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up vote
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down vote
favorite
Recently I have been trying to compile CP/M 2.2 from source. When I try to assemble, everything works except for the instructions ADD A,M and SBC A,M, which the assembler returns a syntax error. I have not done much Z80, so I am a little confused on what the source code means by M?
I am using the CP/M 2.2 Source (Z80 Mnemonics) from the Unofficial CP/M Website.
z80 assembly cp-m
edited 3 hours ago
Alex Hajnal
3,45031632
asked 5 hours ago
tergav17
61
New contributor
tergav17 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Recently I have been trying to compile CP/M 2.2 from source. When I try to assemble, everything works except for the instructions ADD A,M and SBC A,M, which the assembler returns a syntax error. I have not done much Z80, so I am a little confused on what the source code means by M?
I am using the CP/M 2.2 Source (Z80 Mnemonics) from the Unofficial CP/M Website.
z80 assembly cp-m
z80 assembly cp-m
edited 3 hours ago
Alex Hajnal
3,45031632
asked 5 hours ago
tergav17
61
New contributor
tergav17 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Alex Hajnal
3,45031632
asked 5 hours ago
tergav17
61
New contributor
tergav17 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 3 hours ago
Alex Hajnal
3,45031632
edited 3 hours ago
Alex Hajnal
3,45031632
edited 3 hours ago
Alex Hajnal
3,45031632
3,45031632
asked 5 hours ago
tergav17
61
New contributor
tergav17 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
tergav17
61
asked 5 hours ago
tergav17
61
61
New contributor
tergav17 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
tergav17 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
tergav17 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
It might be helpful if you link the source in question and explain where you acquired it.
– Raffzahn
5 hours ago
add a comment |
It might be helpful if you link the source in question and explain where you acquired it.
– Raffzahn
5 hours ago
It might be helpful if you link the source in question and explain where you acquired it.
– Raffzahn
5 hours ago
It might be helpful if you link the source in question and explain where you acquired it.
– Raffzahn
5 hours ago
add a comment |
2 Answers
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In 8080 Assembler M is the memory referenced to by HL.
Depending on the assembler used this would be written as
ADD M(Original Intel 8080 syntax) or
ADD A,M(Later Intel syntax as used for example by CP/M's own ASM (*1))
The Z80 assembler equivalent would be
ADD A,(HL)(Zilog notation)
Are you sure the source you're compiling made for the Z80 (and a Z80 assembler) at all?
CP/M is by default writen in 8080 Assembly, not Z80 or any other substitute. It would make sense that the source you got is meant to be compiled with ASM, as this was the default assembler for CP/M. It would be unusual if it's formated for any later (Z80) assembler.
Wiki got some condensed remarks about the changes Zilog made to the 8080 Assembly Syntax, like the usage of full register as you might have expected in this example. DR's ASM was an inbetween product, adhering (mostly) to Intel syntax while supporting the Z80 as well.
*1 - Later assemblers where often able to compile 8080 as well as Z80 but using 8080 notaton.
answered 5 hours ago
Raffzahn
44.5k5102179
add a comment |
up vote
1
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Depending on the compiler you use, you may need to explicitly state using preprocessor directives that you use 8080 commands. M80 uses .z80 and .8080 to instruct using respective instruction set mnemonics. See https://www.classic-computers.org.nz/system-80/software-manuals/manuals-Macro-80-Assembler.pdf page 20 (16 on the document's page).
If it will not compile in 8080 mode (due to Z80 directives in the source), then it means that sources are altered.The easiest way is to replace M with (HL), but I would not be surprised if resulting executable will not work properly at all.
answered 18 mins ago
Anonymous
83316
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
2
down vote
In 8080 Assembler M is the memory referenced to by HL.
Depending on the assembler used this would be written as
ADD M(Original Intel 8080 syntax) or
ADD A,M(Later Intel syntax as used for example by CP/M's own ASM (*1))
The Z80 assembler equivalent would be
ADD A,(HL)(Zilog notation)
Are you sure the source you're compiling made for the Z80 (and a Z80 assembler) at all?
CP/M is by default writen in 8080 Assembly, not Z80 or any other substitute. It would make sense that the source you got is meant to be compiled with ASM, as this was the default assembler for CP/M. It would be unusual if it's formated for any later (Z80) assembler.
Wiki got some condensed remarks about the changes Zilog made to the 8080 Assembly Syntax, like the usage of full register as you might have expected in this example. DR's ASM was an inbetween product, adhering (mostly) to Intel syntax while supporting the Z80 as well.
*1 - Later assemblers where often able to compile 8080 as well as Z80 but using 8080 notaton.
answered 5 hours ago
Raffzahn
44.5k5102179
add a comment |
up vote
2
down vote
In 8080 Assembler M is the memory referenced to by HL.
Depending on the assembler used this would be written as
ADD M(Original Intel 8080 syntax) or
ADD A,M(Later Intel syntax as used for example by CP/M's own ASM (*1))
The Z80 assembler equivalent would be
ADD A,(HL)(Zilog notation)
Are you sure the source you're compiling made for the Z80 (and a Z80 assembler) at all?
CP/M is by default writen in 8080 Assembly, not Z80 or any other substitute. It would make sense that the source you got is meant to be compiled with ASM, as this was the default assembler for CP/M. It would be unusual if it's formated for any later (Z80) assembler.
Wiki got some condensed remarks about the changes Zilog made to the 8080 Assembly Syntax, like the usage of full register as you might have expected in this example. DR's ASM was an inbetween product, adhering (mostly) to Intel syntax while supporting the Z80 as well.
*1 - Later assemblers where often able to compile 8080 as well as Z80 but using 8080 notaton.
answered 5 hours ago
Raffzahn
44.5k5102179
add a comment |
up vote
2
down vote
up vote
2
down vote
In 8080 Assembler M is the memory referenced to by HL.
Depending on the assembler used this would be written as
ADD M(Original Intel 8080 syntax) or
ADD A,M(Later Intel syntax as used for example by CP/M's own ASM (*1))
The Z80 assembler equivalent would be
ADD A,(HL)(Zilog notation)
Are you sure the source you're compiling made for the Z80 (and a Z80 assembler) at all?
CP/M is by default writen in 8080 Assembly, not Z80 or any other substitute. It would make sense that the source you got is meant to be compiled with ASM, as this was the default assembler for CP/M. It would be unusual if it's formated for any later (Z80) assembler.
Wiki got some condensed remarks about the changes Zilog made to the 8080 Assembly Syntax, like the usage of full register as you might have expected in this example. DR's ASM was an inbetween product, adhering (mostly) to Intel syntax while supporting the Z80 as well.
*1 - Later assemblers where often able to compile 8080 as well as Z80 but using 8080 notaton.
answered 5 hours ago
Raffzahn
44.5k5102179
In 8080 Assembler M is the memory referenced to by HL.
Depending on the assembler used this would be written as
ADD M(Original Intel 8080 syntax) or
ADD A,M(Later Intel syntax as used for example by CP/M's own ASM (*1))
The Z80 assembler equivalent would be
ADD A,(HL)(Zilog notation)
Are you sure the source you're compiling made for the Z80 (and a Z80 assembler) at all?
CP/M is by default writen in 8080 Assembly, not Z80 or any other substitute. It would make sense that the source you got is meant to be compiled with ASM, as this was the default assembler for CP/M. It would be unusual if it's formated for any later (Z80) assembler.
Wiki got some condensed remarks about the changes Zilog made to the 8080 Assembly Syntax, like the usage of full register as you might have expected in this example. DR's ASM was an inbetween product, adhering (mostly) to Intel syntax while supporting the Z80 as well.
*1 - Later assemblers where often able to compile 8080 as well as Z80 but using 8080 notaton.
answered 5 hours ago
Raffzahn
44.5k5102179
edited 4 hours ago
answered 5 hours ago
Raffzahn
44.5k5102179
answered 5 hours ago
Raffzahn
44.5k5102179
answered 5 hours ago
Raffzahn
44.5k5102179
44.5k5102179
add a comment |
add a comment |
up vote
1
down vote
Depending on the compiler you use, you may need to explicitly state using preprocessor directives that you use 8080 commands. M80 uses .z80 and .8080 to instruct using respective instruction set mnemonics. See https://www.classic-computers.org.nz/system-80/software-manuals/manuals-Macro-80-Assembler.pdf page 20 (16 on the document's page).
If it will not compile in 8080 mode (due to Z80 directives in the source), then it means that sources are altered.The easiest way is to replace M with (HL), but I would not be surprised if resulting executable will not work properly at all.
answered 18 mins ago
Anonymous
83316
add a comment |
up vote
1
down vote
Depending on the compiler you use, you may need to explicitly state using preprocessor directives that you use 8080 commands. M80 uses .z80 and .8080 to instruct using respective instruction set mnemonics. See https://www.classic-computers.org.nz/system-80/software-manuals/manuals-Macro-80-Assembler.pdf page 20 (16 on the document's page).
If it will not compile in 8080 mode (due to Z80 directives in the source), then it means that sources are altered.The easiest way is to replace M with (HL), but I would not be surprised if resulting executable will not work properly at all.
answered 18 mins ago
Anonymous
83316
add a comment |
up vote
1
down vote
up vote
1
down vote
Depending on the compiler you use, you may need to explicitly state using preprocessor directives that you use 8080 commands. M80 uses .z80 and .8080 to instruct using respective instruction set mnemonics. See https://www.classic-computers.org.nz/system-80/software-manuals/manuals-Macro-80-Assembler.pdf page 20 (16 on the document's page).
If it will not compile in 8080 mode (due to Z80 directives in the source), then it means that sources are altered.The easiest way is to replace M with (HL), but I would not be surprised if resulting executable will not work properly at all.
answered 18 mins ago
Anonymous
83316
Depending on the compiler you use, you may need to explicitly state using preprocessor directives that you use 8080 commands. M80 uses .z80 and .8080 to instruct using respective instruction set mnemonics. See https://www.classic-computers.org.nz/system-80/software-manuals/manuals-Macro-80-Assembler.pdf page 20 (16 on the document's page).
If it will not compile in 8080 mode (due to Z80 directives in the source), then it means that sources are altered.The easiest way is to replace M with (HL), but I would not be surprised if resulting executable will not work properly at all.
answered 18 mins ago
Anonymous
83316
answered 18 mins ago
Anonymous
83316
answered 18 mins ago
Anonymous
83316
answered 18 mins ago
Anonymous
83316
83316
add a comment |
add a comment |
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– Raffzahn
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