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What was the reason for the ZX Spectrum's display bitmap layout?
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up vote
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I had a ZX Spectrum 48KB, whose display was a 256*192 monochrome bitmap (plus separate per-block color attributes). I remember the display bitmaps scan lines were laid out in memory in a weird way - This was evident when trying to render to the screen via poke, and also when loading games splash screens from tapes.
Why was this layout chosen?
Layout:
- Screen was divided into 3 parts - top/middle/bottom
- In each part, the memory would render the first lines in every 8-line "text line", then the second lines etc.
For example, the layout of the top third would be:
Memory line 0 - Scan line 0
Memory line 1 - Scan line 8
...
Memory line 7 - Scan line 56
Memory line 8 - Scan line 1
Memory line 9 - Scan line 9
...
Memory line 15 - Scan line 57
(skip...)
Memory line 56 - Scan line 7
Memory line 57 - Scan line 15
...
Memory line 63 - Scan line 63
zx-spectrum
asked Aug 14 at 10:41
Jonathan
1734
add a comment |Â
up vote
14
down vote
favorite
I had a ZX Spectrum 48KB, whose display was a 256*192 monochrome bitmap (plus separate per-block color attributes). I remember the display bitmaps scan lines were laid out in memory in a weird way - This was evident when trying to render to the screen via poke, and also when loading games splash screens from tapes.
Why was this layout chosen?
Layout:
- Screen was divided into 3 parts - top/middle/bottom
- In each part, the memory would render the first lines in every 8-line "text line", then the second lines etc.
For example, the layout of the top third would be:
Memory line 0 - Scan line 0
Memory line 1 - Scan line 8
...
Memory line 7 - Scan line 56
Memory line 8 - Scan line 1
Memory line 9 - Scan line 9
...
Memory line 15 - Scan line 57
(skip...)
Memory line 56 - Scan line 7
Memory line 57 - Scan line 15
...
Memory line 63 - Scan line 63
zx-spectrum
asked Aug 14 at 10:41
Jonathan
1734
This book tells you everything you want to know zxdesign.info/book
– Wilson
Aug 14 at 10:50
1
@Wilson Perhaps you could make an answer from that.
– wizzwizz4♦
Aug 14 at 11:06
@wizzwizz4 yep, it's coming this evening. Unless someone beats me to it.
– Wilson
Aug 14 at 11:07
See also the answers to this question: What format is the (Timex) Sinclair ZX Spectrum SCREEN$/.SCR file
– berendi
Aug 15 at 6:08
add a comment |Â
up vote
14
down vote
favorite
up vote
14
down vote
favorite
I had a ZX Spectrum 48KB, whose display was a 256*192 monochrome bitmap (plus separate per-block color attributes). I remember the display bitmaps scan lines were laid out in memory in a weird way - This was evident when trying to render to the screen via poke, and also when loading games splash screens from tapes.
Why was this layout chosen?
Layout:
- Screen was divided into 3 parts - top/middle/bottom
- In each part, the memory would render the first lines in every 8-line "text line", then the second lines etc.
For example, the layout of the top third would be:
Memory line 0 - Scan line 0
Memory line 1 - Scan line 8
...
Memory line 7 - Scan line 56
Memory line 8 - Scan line 1
Memory line 9 - Scan line 9
...
Memory line 15 - Scan line 57
(skip...)
Memory line 56 - Scan line 7
Memory line 57 - Scan line 15
...
Memory line 63 - Scan line 63
zx-spectrum
asked Aug 14 at 10:41
Jonathan
1734
I had a ZX Spectrum 48KB, whose display was a 256*192 monochrome bitmap (plus separate per-block color attributes). I remember the display bitmaps scan lines were laid out in memory in a weird way - This was evident when trying to render to the screen via poke, and also when loading games splash screens from tapes.
Why was this layout chosen?
Layout:
- Screen was divided into 3 parts - top/middle/bottom
- In each part, the memory would render the first lines in every 8-line "text line", then the second lines etc.
For example, the layout of the top third would be:
Memory line 0 - Scan line 0
Memory line 1 - Scan line 8
...
Memory line 7 - Scan line 56
Memory line 8 - Scan line 1
Memory line 9 - Scan line 9
...
Memory line 15 - Scan line 57
(skip...)
Memory line 56 - Scan line 7
Memory line 57 - Scan line 15
...
Memory line 63 - Scan line 63
zx-spectrum
asked Aug 14 at 10:41
Jonathan
1734
asked Aug 14 at 10:41
Jonathan
1734
asked Aug 14 at 10:41
Jonathan
1734
asked Aug 14 at 10:41
Jonathan
1734
1734
This book tells you everything you want to know zxdesign.info/book
– Wilson
Aug 14 at 10:50
1
@Wilson Perhaps you could make an answer from that.
– wizzwizz4♦
Aug 14 at 11:06
@wizzwizz4 yep, it's coming this evening. Unless someone beats me to it.
– Wilson
Aug 14 at 11:07
See also the answers to this question: What format is the (Timex) Sinclair ZX Spectrum SCREEN$/.SCR file
– berendi
Aug 15 at 6:08
add a comment |Â
This book tells you everything you want to know zxdesign.info/book
– Wilson
Aug 14 at 10:50
1
@Wilson Perhaps you could make an answer from that.
– wizzwizz4♦
Aug 14 at 11:06
@wizzwizz4 yep, it's coming this evening. Unless someone beats me to it.
– Wilson
Aug 14 at 11:07
See also the answers to this question: What format is the (Timex) Sinclair ZX Spectrum SCREEN$/.SCR file
– berendi
Aug 15 at 6:08
This book tells you everything you want to know zxdesign.info/book
– Wilson
Aug 14 at 10:50
This book tells you everything you want to know zxdesign.info/book
– Wilson
Aug 14 at 10:50
1
1
@Wilson Perhaps you could make an answer from that.
– wizzwizz4♦
Aug 14 at 11:06
@Wilson Perhaps you could make an answer from that.
– wizzwizz4♦
Aug 14 at 11:06
@wizzwizz4 yep, it's coming this evening. Unless someone beats me to it.
– Wilson
Aug 14 at 11:07
@wizzwizz4 yep, it's coming this evening. Unless someone beats me to it.
– Wilson
Aug 14 at 11:07
See also the answers to this question: What format is the (Timex) Sinclair ZX Spectrum SCREEN$/.SCR file
– berendi
Aug 15 at 6:08
See also the answers to this question: What format is the (Timex) Sinclair ZX Spectrum SCREEN$/.SCR file
– berendi
Aug 15 at 6:08
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
17
down vote
accepted
This is a (for the moment) a short answer:
The Spectrum was engineered with a character oriented display, as Sinclair wanted people to use it for business, not for games, so the screen is arranged so displaying a character (8x8 pixel character) is fast. Given a character position, each scan is separated 256 bytes from the previous one, so to draw a character you will use a 16 bit register as pointer to the display memory and increment the most significant byte to advance one scan. 8 bit increments are fast (4 clocks). A linear display address would have forced each scan to be separated 32 bytes from the previous one, thus making the display of a character a bit slower (the ADD instruction only works with A register, and with an inmediate operand it takes 7 clocks).
To get 256 bytes from one scan to the next one, the display must be divided into three thirds, each one using 8 character rows. At 32 characters per row, and being each character a grid of 8x8 pixels, this allows the desired 256 byte separation.The ULA uses DRAM page read so it reads two bytes faster than two separate reads of one byte each. The ULA needs for each scan of 8 pixels, one byte of bitmap and one byte of attributes. To be able to use page read, addresses must share either the most significant 7 bits, or the least significant 7 bits. If the ULA would were engineered so the addresses share the most significant 7 bits, that would mean that page reads would read consecutive addresses, and that would mean that the display would have to be arranged so bitmap and attributes would have to be interleaved, making impossible to have 256 bytes between two adjacent scans in the same char.
So page mode read addresses share the 7 least significant bits and change the 7 most significant ones, thus allowing the attribute zone to be completely separated from the bitmap zone.
answered Aug 14 at 11:55
mcleod_ideafix
11.1k3772
Ula MUST share LSB 7 bits (that must be RAS addresses on DRAM chips) since Z80 refresh circuitry is based on the same principle: during Z80 refresh cycles, the contents of R register is put on A6..A0 address pins.
– lvd
Aug 14 at 16:56
Another view on the requirements to have such a puzzled screen: retrocomputing.stackexchange.com/a/4858/4692 In short, without page-mode access, Z80 would have been stopped for the complete duration of line fetch. Due to ingenious page-mode access that ULA makes to video DRAM, Z80 access to video memory on ZX is still possible, with the famous 6-5-4-3-2-1-0-0 pattern.
– lvd
Aug 14 at 17:00
ADD HL, [BC/DE/HL/SP]slightly disagree with the reasoning in your first bullet point, but not the conclusion. There's no immediate 16-bit add, so you'd lose a register pair to store the32constant, and at 11 cycles it's still slower than a simpleINC H.
– Tommy
Aug 14 at 17:49
@lvd - even with a linear layout, the ULA could have used page mode to optimize accesses, but it would have needed to read 16 pixels ahead rather than just 8 each time, which would have made the required circuitry larger and the ULA potentially more expensive. Alternatively, the RAM chips could have been organised with their address lines mapped into a different order than the one the processor expects; this would have worked (and, I believe, was the solution used in the Apple II for refresh, which is a similar problem).
– Jules
Aug 14 at 18:37
@Jules -- could you please give more info on how linear or another different layouts would be possible within the given DRAM restrictions?
– lvd
Aug 14 at 20:02
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
17
down vote
accepted
This is a (for the moment) a short answer:
The Spectrum was engineered with a character oriented display, as Sinclair wanted people to use it for business, not for games, so the screen is arranged so displaying a character (8x8 pixel character) is fast. Given a character position, each scan is separated 256 bytes from the previous one, so to draw a character you will use a 16 bit register as pointer to the display memory and increment the most significant byte to advance one scan. 8 bit increments are fast (4 clocks). A linear display address would have forced each scan to be separated 32 bytes from the previous one, thus making the display of a character a bit slower (the ADD instruction only works with A register, and with an inmediate operand it takes 7 clocks).
To get 256 bytes from one scan to the next one, the display must be divided into three thirds, each one using 8 character rows. At 32 characters per row, and being each character a grid of 8x8 pixels, this allows the desired 256 byte separation.The ULA uses DRAM page read so it reads two bytes faster than two separate reads of one byte each. The ULA needs for each scan of 8 pixels, one byte of bitmap and one byte of attributes. To be able to use page read, addresses must share either the most significant 7 bits, or the least significant 7 bits. If the ULA would were engineered so the addresses share the most significant 7 bits, that would mean that page reads would read consecutive addresses, and that would mean that the display would have to be arranged so bitmap and attributes would have to be interleaved, making impossible to have 256 bytes between two adjacent scans in the same char.
So page mode read addresses share the 7 least significant bits and change the 7 most significant ones, thus allowing the attribute zone to be completely separated from the bitmap zone.
answered Aug 14 at 11:55
mcleod_ideafix
11.1k3772
Ula MUST share LSB 7 bits (that must be RAS addresses on DRAM chips) since Z80 refresh circuitry is based on the same principle: during Z80 refresh cycles, the contents of R register is put on A6..A0 address pins.
– lvd
Aug 14 at 16:56
Another view on the requirements to have such a puzzled screen: retrocomputing.stackexchange.com/a/4858/4692 In short, without page-mode access, Z80 would have been stopped for the complete duration of line fetch. Due to ingenious page-mode access that ULA makes to video DRAM, Z80 access to video memory on ZX is still possible, with the famous 6-5-4-3-2-1-0-0 pattern.
– lvd
Aug 14 at 17:00
ADD HL, [BC/DE/HL/SP]slightly disagree with the reasoning in your first bullet point, but not the conclusion. There's no immediate 16-bit add, so you'd lose a register pair to store the32constant, and at 11 cycles it's still slower than a simpleINC H.
– Tommy
Aug 14 at 17:49
@lvd - even with a linear layout, the ULA could have used page mode to optimize accesses, but it would have needed to read 16 pixels ahead rather than just 8 each time, which would have made the required circuitry larger and the ULA potentially more expensive. Alternatively, the RAM chips could have been organised with their address lines mapped into a different order than the one the processor expects; this would have worked (and, I believe, was the solution used in the Apple II for refresh, which is a similar problem).
– Jules
Aug 14 at 18:37
@Jules -- could you please give more info on how linear or another different layouts would be possible within the given DRAM restrictions?
– lvd
Aug 14 at 20:02
 |Â
show 1 more comment
up vote
17
down vote
accepted
This is a (for the moment) a short answer:
The Spectrum was engineered with a character oriented display, as Sinclair wanted people to use it for business, not for games, so the screen is arranged so displaying a character (8x8 pixel character) is fast. Given a character position, each scan is separated 256 bytes from the previous one, so to draw a character you will use a 16 bit register as pointer to the display memory and increment the most significant byte to advance one scan. 8 bit increments are fast (4 clocks). A linear display address would have forced each scan to be separated 32 bytes from the previous one, thus making the display of a character a bit slower (the ADD instruction only works with A register, and with an inmediate operand it takes 7 clocks).
To get 256 bytes from one scan to the next one, the display must be divided into three thirds, each one using 8 character rows. At 32 characters per row, and being each character a grid of 8x8 pixels, this allows the desired 256 byte separation.The ULA uses DRAM page read so it reads two bytes faster than two separate reads of one byte each. The ULA needs for each scan of 8 pixels, one byte of bitmap and one byte of attributes. To be able to use page read, addresses must share either the most significant 7 bits, or the least significant 7 bits. If the ULA would were engineered so the addresses share the most significant 7 bits, that would mean that page reads would read consecutive addresses, and that would mean that the display would have to be arranged so bitmap and attributes would have to be interleaved, making impossible to have 256 bytes between two adjacent scans in the same char.
So page mode read addresses share the 7 least significant bits and change the 7 most significant ones, thus allowing the attribute zone to be completely separated from the bitmap zone.
answered Aug 14 at 11:55
mcleod_ideafix
11.1k3772
Ula MUST share LSB 7 bits (that must be RAS addresses on DRAM chips) since Z80 refresh circuitry is based on the same principle: during Z80 refresh cycles, the contents of R register is put on A6..A0 address pins.
– lvd
Aug 14 at 16:56
Another view on the requirements to have such a puzzled screen: retrocomputing.stackexchange.com/a/4858/4692 In short, without page-mode access, Z80 would have been stopped for the complete duration of line fetch. Due to ingenious page-mode access that ULA makes to video DRAM, Z80 access to video memory on ZX is still possible, with the famous 6-5-4-3-2-1-0-0 pattern.
– lvd
Aug 14 at 17:00
ADD HL, [BC/DE/HL/SP]slightly disagree with the reasoning in your first bullet point, but not the conclusion. There's no immediate 16-bit add, so you'd lose a register pair to store the32constant, and at 11 cycles it's still slower than a simpleINC H.
– Tommy
Aug 14 at 17:49
@lvd - even with a linear layout, the ULA could have used page mode to optimize accesses, but it would have needed to read 16 pixels ahead rather than just 8 each time, which would have made the required circuitry larger and the ULA potentially more expensive. Alternatively, the RAM chips could have been organised with their address lines mapped into a different order than the one the processor expects; this would have worked (and, I believe, was the solution used in the Apple II for refresh, which is a similar problem).
– Jules
Aug 14 at 18:37
@Jules -- could you please give more info on how linear or another different layouts would be possible within the given DRAM restrictions?
– lvd
Aug 14 at 20:02
 |Â
show 1 more comment
up vote
17
down vote
accepted
up vote
17
down vote
accepted
This is a (for the moment) a short answer:
The Spectrum was engineered with a character oriented display, as Sinclair wanted people to use it for business, not for games, so the screen is arranged so displaying a character (8x8 pixel character) is fast. Given a character position, each scan is separated 256 bytes from the previous one, so to draw a character you will use a 16 bit register as pointer to the display memory and increment the most significant byte to advance one scan. 8 bit increments are fast (4 clocks). A linear display address would have forced each scan to be separated 32 bytes from the previous one, thus making the display of a character a bit slower (the ADD instruction only works with A register, and with an inmediate operand it takes 7 clocks).
To get 256 bytes from one scan to the next one, the display must be divided into three thirds, each one using 8 character rows. At 32 characters per row, and being each character a grid of 8x8 pixels, this allows the desired 256 byte separation.The ULA uses DRAM page read so it reads two bytes faster than two separate reads of one byte each. The ULA needs for each scan of 8 pixels, one byte of bitmap and one byte of attributes. To be able to use page read, addresses must share either the most significant 7 bits, or the least significant 7 bits. If the ULA would were engineered so the addresses share the most significant 7 bits, that would mean that page reads would read consecutive addresses, and that would mean that the display would have to be arranged so bitmap and attributes would have to be interleaved, making impossible to have 256 bytes between two adjacent scans in the same char.
So page mode read addresses share the 7 least significant bits and change the 7 most significant ones, thus allowing the attribute zone to be completely separated from the bitmap zone.
answered Aug 14 at 11:55
mcleod_ideafix
11.1k3772
This is a (for the moment) a short answer:
The Spectrum was engineered with a character oriented display, as Sinclair wanted people to use it for business, not for games, so the screen is arranged so displaying a character (8x8 pixel character) is fast. Given a character position, each scan is separated 256 bytes from the previous one, so to draw a character you will use a 16 bit register as pointer to the display memory and increment the most significant byte to advance one scan. 8 bit increments are fast (4 clocks). A linear display address would have forced each scan to be separated 32 bytes from the previous one, thus making the display of a character a bit slower (the ADD instruction only works with A register, and with an inmediate operand it takes 7 clocks).
To get 256 bytes from one scan to the next one, the display must be divided into three thirds, each one using 8 character rows. At 32 characters per row, and being each character a grid of 8x8 pixels, this allows the desired 256 byte separation.The ULA uses DRAM page read so it reads two bytes faster than two separate reads of one byte each. The ULA needs for each scan of 8 pixels, one byte of bitmap and one byte of attributes. To be able to use page read, addresses must share either the most significant 7 bits, or the least significant 7 bits. If the ULA would were engineered so the addresses share the most significant 7 bits, that would mean that page reads would read consecutive addresses, and that would mean that the display would have to be arranged so bitmap and attributes would have to be interleaved, making impossible to have 256 bytes between two adjacent scans in the same char.
So page mode read addresses share the 7 least significant bits and change the 7 most significant ones, thus allowing the attribute zone to be completely separated from the bitmap zone.
answered Aug 14 at 11:55
mcleod_ideafix
11.1k3772
answered Aug 14 at 11:55
mcleod_ideafix
11.1k3772
answered Aug 14 at 11:55
mcleod_ideafix
11.1k3772
answered Aug 14 at 11:55
mcleod_ideafix
11.1k3772
11.1k3772
Ula MUST share LSB 7 bits (that must be RAS addresses on DRAM chips) since Z80 refresh circuitry is based on the same principle: during Z80 refresh cycles, the contents of R register is put on A6..A0 address pins.
– lvd
Aug 14 at 16:56
Another view on the requirements to have such a puzzled screen: retrocomputing.stackexchange.com/a/4858/4692 In short, without page-mode access, Z80 would have been stopped for the complete duration of line fetch. Due to ingenious page-mode access that ULA makes to video DRAM, Z80 access to video memory on ZX is still possible, with the famous 6-5-4-3-2-1-0-0 pattern.
– lvd
Aug 14 at 17:00
ADD HL, [BC/DE/HL/SP]slightly disagree with the reasoning in your first bullet point, but not the conclusion. There's no immediate 16-bit add, so you'd lose a register pair to store the32constant, and at 11 cycles it's still slower than a simpleINC H.
– Tommy
Aug 14 at 17:49
@lvd - even with a linear layout, the ULA could have used page mode to optimize accesses, but it would have needed to read 16 pixels ahead rather than just 8 each time, which would have made the required circuitry larger and the ULA potentially more expensive. Alternatively, the RAM chips could have been organised with their address lines mapped into a different order than the one the processor expects; this would have worked (and, I believe, was the solution used in the Apple II for refresh, which is a similar problem).
– Jules
Aug 14 at 18:37
@Jules -- could you please give more info on how linear or another different layouts would be possible within the given DRAM restrictions?
– lvd
Aug 14 at 20:02
 |Â
show 1 more comment
Ula MUST share LSB 7 bits (that must be RAS addresses on DRAM chips) since Z80 refresh circuitry is based on the same principle: during Z80 refresh cycles, the contents of R register is put on A6..A0 address pins.
– lvd
Aug 14 at 16:56
Another view on the requirements to have such a puzzled screen: retrocomputing.stackexchange.com/a/4858/4692 In short, without page-mode access, Z80 would have been stopped for the complete duration of line fetch. Due to ingenious page-mode access that ULA makes to video DRAM, Z80 access to video memory on ZX is still possible, with the famous 6-5-4-3-2-1-0-0 pattern.
– lvd
Aug 14 at 17:00
ADD HL, [BC/DE/HL/SP]slightly disagree with the reasoning in your first bullet point, but not the conclusion. There's no immediate 16-bit add, so you'd lose a register pair to store the32constant, and at 11 cycles it's still slower than a simpleINC H.
– Tommy
Aug 14 at 17:49
@lvd - even with a linear layout, the ULA could have used page mode to optimize accesses, but it would have needed to read 16 pixels ahead rather than just 8 each time, which would have made the required circuitry larger and the ULA potentially more expensive. Alternatively, the RAM chips could have been organised with their address lines mapped into a different order than the one the processor expects; this would have worked (and, I believe, was the solution used in the Apple II for refresh, which is a similar problem).
– Jules
Aug 14 at 18:37
@Jules -- could you please give more info on how linear or another different layouts would be possible within the given DRAM restrictions?
– lvd
Aug 14 at 20:02
Ula MUST share LSB 7 bits (that must be RAS addresses on DRAM chips) since Z80 refresh circuitry is based on the same principle: during Z80 refresh cycles, the contents of R register is put on A6..A0 address pins.
– lvd
Aug 14 at 16:56
Ula MUST share LSB 7 bits (that must be RAS addresses on DRAM chips) since Z80 refresh circuitry is based on the same principle: during Z80 refresh cycles, the contents of R register is put on A6..A0 address pins.
– lvd
Aug 14 at 16:56
Another view on the requirements to have such a puzzled screen: retrocomputing.stackexchange.com/a/4858/4692 In short, without page-mode access, Z80 would have been stopped for the complete duration of line fetch. Due to ingenious page-mode access that ULA makes to video DRAM, Z80 access to video memory on ZX is still possible, with the famous 6-5-4-3-2-1-0-0 pattern.
– lvd
Aug 14 at 17:00
Another view on the requirements to have such a puzzled screen: retrocomputing.stackexchange.com/a/4858/4692 In short, without page-mode access, Z80 would have been stopped for the complete duration of line fetch. Due to ingenious page-mode access that ULA makes to video DRAM, Z80 access to video memory on ZX is still possible, with the famous 6-5-4-3-2-1-0-0 pattern.
– lvd
Aug 14 at 17:00
ADD HL, [BC/DE/HL/SP] slightly disagree with the reasoning in your first bullet point, but not the conclusion. There's no immediate 16-bit add, so you'd lose a register pair to store the 32 constant, and at 11 cycles it's still slower than a simple INC H.– Tommy
Aug 14 at 17:49
ADD HL, [BC/DE/HL/SP] slightly disagree with the reasoning in your first bullet point, but not the conclusion. There's no immediate 16-bit add, so you'd lose a register pair to store the 32 constant, and at 11 cycles it's still slower than a simple INC H.– Tommy
Aug 14 at 17:49
@lvd - even with a linear layout, the ULA could have used page mode to optimize accesses, but it would have needed to read 16 pixels ahead rather than just 8 each time, which would have made the required circuitry larger and the ULA potentially more expensive. Alternatively, the RAM chips could have been organised with their address lines mapped into a different order than the one the processor expects; this would have worked (and, I believe, was the solution used in the Apple II for refresh, which is a similar problem).
– Jules
Aug 14 at 18:37
@lvd - even with a linear layout, the ULA could have used page mode to optimize accesses, but it would have needed to read 16 pixels ahead rather than just 8 each time, which would have made the required circuitry larger and the ULA potentially more expensive. Alternatively, the RAM chips could have been organised with their address lines mapped into a different order than the one the processor expects; this would have worked (and, I believe, was the solution used in the Apple II for refresh, which is a similar problem).
– Jules
Aug 14 at 18:37
@Jules -- could you please give more info on how linear or another different layouts would be possible within the given DRAM restrictions?
– lvd
Aug 14 at 20:02
@Jules -- could you please give more info on how linear or another different layouts would be possible within the given DRAM restrictions?
– lvd
Aug 14 at 20:02
 |Â
show 1 more comment
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This book tells you everything you want to know zxdesign.info/book
– Wilson
Aug 14 at 10:50
1
@Wilson Perhaps you could make an answer from that.
– wizzwizz4♦
Aug 14 at 11:06
@wizzwizz4 yep, it's coming this evening. Unless someone beats me to it.
– Wilson
Aug 14 at 11:07
See also the answers to this question: What format is the (Timex) Sinclair ZX Spectrum SCREEN$/.SCR file
– berendi
Aug 15 at 6:08