Short, short questions
I'm about to finish my lunch break, sorry if I sound slightly short... 1) is there any pattern or logic to the values placed on the bus during interrupts? I guess an equivalent question is: what realistic options do I have on the Sam for catching and processing interrupts? 2) are there any non-obvious tricks for fast access to a table containing 16bit words, indexed by an 11bit (signed) integer? At the minute I'm essentially doing: [stuff to work out offset into table in hl] ld bc, address of middle of table - which is aligned to a two-byte boundary add hl, hl add hl, bc ld e, (hl) inc l ld d, (hl)
Re: Short, short questions
On Tue, May 20, 2008 at 02:03:57PM +0100, Thomas Harte wrote: I'm about to finish my lunch break, sorry if I sound slightly short... 1) is there any pattern or logic to the values placed on the bus during interrupts? I guess an equivalent question is: what realistic options do I have on the Sam for catching and processing interrupts? I don't think the bus value is ever constructed in a useful way. The usual strategies are to use mode 1, or to use mode 2 with a 257-byte table all containing the same byte. 2) are there any non-obvious tricks for fast access to a table containing 16bit words, indexed by an 11bit (signed) integer? At the minute I'm essentially doing: [stuff to work out offset into table in hl] ld bc, address of middle of table - which is aligned to a two-byte boundary add hl, hl add hl, bc ld e, (hl) inc l ld d, (hl) If you can align your table onto a 256 byte boundary, you can save a few cycles by adding only the top byte of the table offset. You can also double a number more quickly if you have an excuse to put some of it in the A register: ld a,h 4 sla h 8 rla4 add table/256 8 ld h,a 4 = 28 instead of ld bc, table 12 add hl, hl 16 add hl, bc 16 = 44 HTH, Andrew -- --- Andrew Collier http://www.intensity.org.uk/ --- -- r2+ T4* cSEL dMS hEn/CBBL A4 S+*++ C$++L/mP W- a-- Vh+seT+ (Cantab) 1.1.4
Re: Short, short questions
On Tue, May 20, 2008 at 03:22:54PM +0100, Andrew Collier wrote: correction: ld a,h 4 sla l 8 rla4 add table/256 8 ld h,a 4 = 28 You can also shave a little more time if you're willing to rearrange the table: Instead of word pairs (low byte, high byte, low byte, high byte) you could have alternating lines of 256 low bytes, 256 high bytes. To use that, double the high byte of the address but don't change the low byte (in other words, don't run the 'sla l' at all saving 8 t-states) and then, when reading DE from the table, increment H instead of incrementing L to get the high byte corresponding to the selected low byte. HTH, Andrew -- --- Andrew Collier http://www.intensity.org.uk/ --- -- r2+ T4* cSEL dMS hEn/CBBL A4 S+*++ C$++L/mP W- a-- Vh+seT+ (Cantab) 1.1.4
Re: Short, short questions
While thinking about it during the afternoon, I have decided that since this table is absolutely fundamental to my program (it's part of the table based 2.8x2.8 multiply), I'm just going to centre it on address 0. Then I can cut the ld bc and the add hl, bc. I'm happy to rearrange my table, so I guess the comparison is between: add hl, hl ld e, (hl) inc l ld d, (hl) and: sla h ld e, (hl) inc h ld d, (hl) Obviously the second is faster — but I'm curious about your cycle counts. All the z80 documentation I have lists add hl, ss as 11 t- states. Why have you turned that into 16? It looks from your other calculations that you're just rounding to the next multiple of 4 (which may be a good rule of thumb for the Sam, I don't know, I've just been letting Sim Coupe work it out), so why isn't it 12? Anyway, with moving the table to 0 and rearranging it I have a reasonably accurate 2.8 x 2.8 multiply that operates in just 109 paper cycles, or between 152 and 268 Sim Coupe cycles. Annoyingly I currently have the screen in the low 32kb and my program in the high 32kb so there's a whole bunch of things to change before I can actually see what overall effect that has on my framerate versus my current 200 to 304 Sim Coupe cycles... On 20 May 2008, at 17:28, Andrew Collier wrote: On Tue, May 20, 2008 at 03:22:54PM +0100, Andrew Collier wrote: correction: ld a,h 4 sla l 8 rla4 add table/256 8 ld h,a 4 = 28 You can also shave a little more time if you're willing to rearrange the table: Instead of word pairs (low byte, high byte, low byte, high byte) you could have alternating lines of 256 low bytes, 256 high bytes. To use that, double the high byte of the address but don't change the low byte (in other words, don't run the 'sla l' at all saving 8 t-states) and then, when reading DE from the table, increment H instead of incrementing L to get the high byte corresponding to the selected low byte. HTH, Andrew -- --- Andrew Collier http://www.intensity.org.uk/ --- -- r2+ T4* cSEL dMS hEn/CBBL A4 S+*++ C$++L/mP W- a-- Vh+seT+ (Cantab) 1.1.4
Re: Short, short questions
On Tue, May 20, 2008 at 07:20:48PM +0100, Thomas Harte wrote: Obviously the second is faster — but I'm curious about your cycle counts. All the z80 documentation I have lists add hl, ss as 11 t- states. Why have you turned that into 16? By mistake :) I'm working from google, and misread a table. You're right, it should be 12. (Whn I get home I'll check there aren't any other details I missed). There's a very comprehensive article in Based On An Idea about Sam timings; in almost all cases you can consider the real timing to be the Z80 timings rounded up to multiples of 4 (and then doubled if you're in the screen area - because there's little way top avoid this effect, I find it simpler to think of timings in terms of a t-state whose length changes). Then as the second layer of approximation, there are some instructions which don't access memory so much, and take less than double time in the screen area. In effect you recover 4 of the variable t-states if you happen to be in the screen. The article lists there with a * for each 4 t-states you can save. INC rr is one (there are more, I can't remember right now) which takes 8* t-states. Andrew -- --- Andrew Collier http://www.intensity.org.uk/ --- -- r2+ T4* cSEL dMS hEn/CBBL A4 S+*++ C$++L/mP W- a-- Vh+seT+ (Cantab) 1.1.4
Re: Short, short questions
1) Normally the value is FF and it looks pretty stable too considering PRO-DOS uses IM 2 without a vector table. So your choices are IM 0/IM 1 both do a RST 38H or IM 2 which 'calls' the adress found at the address pointed by I register (MSB) and the value on the databus (LSB). Like the speccy there may be some hardware connected that changes the default value on the databus so a 257 byte vector table is recomended. 2) If choose to put all the LSB at 32768 followed by the MSB then you could use the following: set 7,h lde,(hl) set 3,h ldd,(hl) Other usefull addresses for the table would be 16384 (use SET 6,H), 8192 (SET 5,H), 4096 (SET 4,H) or 0 (you don't need the first SET instruction at all) Edwin - Original Message - From: Thomas Harte [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 3:03 PM Subject: Short, short questions I'm about to finish my lunch break, sorry if I sound slightly short... 1) is there any pattern or logic to the values placed on the bus during interrupts? I guess an equivalent question is: what realistic options do I have on the Sam for catching and processing interrupts? 2) are there any non-obvious tricks for fast access to a table containing 16bit words, indexed by an 11bit (signed) integer? At the minute I'm essentially doing: [stuff to work out offset into table in hl] ld bc, address of middle of table - which is aligned to a two-byte boundary add hl, hl add hl, bc ld e, (hl) inc l ld d, (hl)
Re: Short, short questions
Mode 2 uses a table with 128 word address but as byte high,byte low not the normal low, high bytes So if you set your org/dump address to ??FF (i.e. ??00-1) and then do DEFWmode2.i,mode2.i so you have 129 words. mode2.i: di pushaf ina,(status.int) . . ei ret - Original Message - From: Andrew Collier [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 3:22 PM Subject: Re: Short, short questions The usual strategies are to use mode 1, or to use mode 2 with a 257-byte table all containing the same byte.
Re: Short, short questions
Surely set would only work if I had an unsigned offset from the beginning of a table? I'm using a signed offset from the middle of a table. On 20 May 2008, at 19:48, Edwin Blink wrote: 1) Normally the value is FF and it looks pretty stable too considering PRO-DOS uses IM 2 without a vector table. So your choices are IM 0/ IM 1 both do a RST 38H or IM 2 which 'calls' the adress found at the address pointed by I register (MSB) and the value on the databus (LSB). Like the speccy there may be some hardware connected that changes the default value on the databus so a 257 byte vector table is recomended. 2) If choose to put all the LSB at 32768 followed by the MSB then you could use the following: set 7,h lde,(hl) set 3,h ldd,(hl) Other usefull addresses for the table would be 16384 (use SET 6,H), 8192 (SET 5,H), 4096 (SET 4,H) or 0 (you don't need the first SET instruction at all) Edwin - Original Message - From: Thomas Harte [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 3:03 PM Subject: Short, short questions I'm about to finish my lunch break, sorry if I sound slightly short... 1) is there any pattern or logic to the values placed on the bus during interrupts? I guess an equivalent question is: what realistic options do I have on the Sam for catching and processing interrupts? 2) are there any non-obvious tricks for fast access to a table containing 16bit words, indexed by an 11bit (signed) integer? At the minute I'm essentially doing: [stuff to work out offset into table in hl] ld bc, address of middle of table - which is aligned to a two-byte boundary add hl, hl add hl, bc ld e, (hl) inc l ld d, (hl)
Re: Short, short questions
From: Thomas Harte Surely set would only work if I had an unsigned offset from the beginning of a table? I'm using a signed offset from the middle of a table. it works the same as ADD HL,HL ADD HL,BC the difference is you store the table entries LSB,MSB so you need the ADD HL,HL First you get the LSB from the table then INC L points to MSB When you store LSB,LSB... for all table enties first and then MSB,MSB... you don't need to do the add as HL already points to table entries LSB to get the MSB you add 2K to get to MSB Or am I getting something wrong ? Because you have your table at address 0 the following would do : LD E,(HL) set 3,H;Add 2K to point to MSB LD D,(HL) takes 24Ts :-) Edwin
Re: Short, short questions
Boy Do I feel silly I missed that ADD HL,BC. Edwin - Original Message - From: Thomas Harte [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 9:53 PM Subject: Re: Short, short questions Surely set would only work if I had an unsigned offset from the beginning of a table? I'm using a signed offset from the middle of a table. On 20 May 2008, at 19:48, Edwin Blink wrote: 1) Normally the value is FF and it looks pretty stable too considering PRO-DOS uses IM 2 without a vector table. So your choices are IM 0/ IM 1 both do a RST 38H or IM 2 which 'calls' the adress found at the address pointed by I register (MSB) and the value on the databus (LSB). Like the speccy there may be some hardware connected that changes the default value on the databus so a 257 byte vector table is recomended. 2) If choose to put all the LSB at 32768 followed by the MSB then you could use the following: set 7,h lde,(hl) set 3,h ldd,(hl) Other usefull addresses for the table would be 16384 (use SET 6,H), 8192 (SET 5,H), 4096 (SET 4,H) or 0 (you don't need the first SET instruction at all) Edwin - Original Message - From: Thomas Harte [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 3:03 PM Subject: Short, short questions I'm about to finish my lunch break, sorry if I sound slightly short... 1) is there any pattern or logic to the values placed on the bus during interrupts? I guess an equivalent question is: what realistic options do I have on the Sam for catching and processing interrupts? 2) are there any non-obvious tricks for fast access to a table containing 16bit words, indexed by an 11bit (signed) integer? At the minute I'm essentially doing: [stuff to work out offset into table in hl] ld bc, address of middle of table - which is aligned to a two-byte boundary add hl, hl add hl, bc ld e, (hl) inc l ld d, (hl)
Re: Short, short questions
Hi, I'm sceptical about this claim. I've never heard anybody say that the vector formed is big-endian - it's just you don't know the byte offset from which the interrupt vector will be fetched. (As Edwin says, it is usually 255 - which is odd so your 1-aligned table will usually work - but I don't know that Sam's hardware guarantees this). So the high byte comes from I, the low byte from the data bus; this forms a 16 bit address which will be incremented once (which is why the table needs 257 bytes, not 256). You could, at least in theory, read the vector address from even or odd overlapping entries, which is why the usual strategy is to pick a vector address whose low and high bytes are the same. The last IM2 interrupt routine I wrote looked something like this: ds ALIGN 256 IM2TABLE: equ $ IM2BYTE:equ im2table/256 IM2TARGETBYTE: equ IM2BYTE+1 for 257, DB IM2TARGETBYTE IM2TARGET: equ 257*IM2TARGETBYTE ds IM2TARGET-$ EX AF,AF' ... Andrew On 20 May 2008, at 21:16, David Brant wrote: Mode 2 uses a table with 128 word address but as byte high,byte low not the normal low, high bytes So if you set your org/dump address to ??FF (i.e. ??00-1) and then do DEFWmode2.i,mode2.i so you have 129 words. mode2.i: di pushaf ina,(status.int) . . ei ret - Original Message - From: Andrew Collier [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 3:22 PM Subject: Re: Short, short questions The usual strategies are to use mode 1, or to use mode 2 with a 257- byte table all containing the same byte. -- --- Andrew Collier http://www.intensity.org.uk/ --- --
Re: Short, short questions
This was based on info from a book called z-80 Workshop manual by E.A Parr. The I register gives the high part of the table and the hardware gives the low part to the table then takes that word for the service routine. So if you start from one byte before the table and use the same address for all entries and over run it by one it will work. My demo of a full scrolling football pitch used this system, which I believe you saw many years a go. Dave - Original Message - From: Andrew Collier [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 9:50 PM Subject: Re: Short, short questions Hi, I'm sceptical about this claim. I've never heard anybody say that the vector formed is big-endian - it's just you don't know the byte offset from which the interrupt vector will be fetched. (As Edwin says, it is usually 255 - which is odd so your 1-aligned table will usually work - but I don't know that Sam's hardware guarantees this). So the high byte comes from I, the low byte from the data bus; this forms a 16 bit address which will be incremented once (which is why the table needs 257 bytes, not 256). You could, at least in theory, read the vector address from even or odd overlapping entries, which is why the usual strategy is to pick a vector address whose low and high bytes are the same. The last IM2 interrupt routine I wrote looked something like this: ds ALIGN 256 IM2TABLE: equ $ IM2BYTE: equ im2table/256 IM2TARGETBYTE: equ IM2BYTE+1 for 257, DB IM2TARGETBYTE IM2TARGET: equ 257*IM2TARGETBYTE ds IM2TARGET-$ EX AF,AF' ... Andrew On 20 May 2008, at 21:16, David Brant wrote: Mode 2 uses a table with 128 word address but as byte high,byte low not the normal low, high bytes So if you set your org/dump address to ??FF (i.e. ??00-1) and then do DEFWmode2.i,mode2.i so you have 129 words. mode2.i: di pushaf ina,(status.int) . . ei ret - Original Message - From: Andrew Collier [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 3:22 PM Subject: Re: Short, short questions The usual strategies are to use mode 1, or to use mode 2 with a 257- byte table all containing the same byte. -- --- Andrew Collier http://www.intensity.org.uk/ --- --
Re: Short, short questions
I've just been looking at my books. Although I can't find the bit that said about swapping to high,low but I'm sure that I did read it somewhere. It does say that the device only gives the bits 1-7 and bit 0 is always 0 giving 128 possible addresses. Dave - Original Message - From: David Brant [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 10:49 PM Subject: Re: Short, short questions This was based on info from a book called z-80 Workshop manual by E.A Parr. The I register gives the high part of the table and the hardware gives the low part to the table then takes that word for the service routine. So if you start from one byte before the table and use the same address for all entries and over run it by one it will work. My demo of a full scrolling football pitch used this system, which I believe you saw many years a go. Dave - Original Message - From: Andrew Collier [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 9:50 PM Subject: Re: Short, short questions Hi, I'm sceptical about this claim. I've never heard anybody say that the vector formed is big-endian - it's just you don't know the byte offset from which the interrupt vector will be fetched. (As Edwin says, it is usually 255 - which is odd so your 1-aligned table will usually work - but I don't know that Sam's hardware guarantees this). So the high byte comes from I, the low byte from the data bus; this forms a 16 bit address which will be incremented once (which is why the table needs 257 bytes, not 256). You could, at least in theory, read the vector address from even or odd overlapping entries, which is why the usual strategy is to pick a vector address whose low and high bytes are the same. The last IM2 interrupt routine I wrote looked something like this: ds ALIGN 256 IM2TABLE: equ $ IM2BYTE: equ im2table/256 IM2TARGETBYTE: equ IM2BYTE+1 for 257, DB IM2TARGETBYTE IM2TARGET: equ 257*IM2TARGETBYTE ds IM2TARGET-$ EX AF,AF' ... Andrew On 20 May 2008, at 21:16, David Brant wrote: Mode 2 uses a table with 128 word address but as byte high,byte low not the normal low, high bytes So if you set your org/dump address to ??FF (i.e. ??00-1) and then do DEFWmode2.i,mode2.i so you have 129 words. mode2.i: di pushaf ina,(status.int) . . ei ret - Original Message - From: Andrew Collier [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 3:22 PM Subject: Re: Short, short questions The usual strategies are to use mode 1, or to use mode 2 with a 257- byte table all containing the same byte. -- --- Andrew Collier http://www.intensity.org.uk/ --- --
Re: Short, short questions
Hi, In a spirit of scientific enquiry, let us perform an experiment by running the code below. On SimCoupe, at least, the border turns bright yellow, meaning it ran the interrupt vector at 0x9a92 (that is, the 256th and 257th bytes of the table). Anybody got real hardware handy? Cheers, Andrew org 32768 dump 32768 di ld a,TABLEADDR/256 ld i,a im 2 ei @: halt jr @ TABLEADDR: equ 8100 ds TABLEADDR - $ for 254, DB 82 DB 8a, 92, 9a BORDER: EQU 254 ds 8282 - $ ld a,0 out (BORDER),a reti ds 828a - $ ld a,1 out (BORDER),a reti ds 8292 - $ ld a,2 out (BORDER),a reti ds 829a - $ ld a,3 out (BORDER),a reti ds 8a82 - $ ld a,4 out (BORDER),a reti ds 8a8a - $ ld a,5 out (BORDER),a reti ds 8a92 - $ ld a,17 out (BORDER),a reti ds 8a9a - $ ld a,7 out (BORDER),a reti ds 9282 - $ ld a,32 out (BORDER),a reti ds 928a - $ ld a,33 out (BORDER),a reti ds 9292 - $ ld a,34 out (BORDER),a reti ds 929a - $ ld a,35 out (BORDER),a reti ds 9a82 - $ ld a,36 out (BORDER),a reti ds 9a8a - $ ld a,37 out (BORDER),a reti ds 9a92 - $ ld a,38 out (BORDER),a reti ds 9a9a - $ ld a,39 out (BORDER),a reti On 21 May 2008, at 00:02, David Brant wrote: I've just been looking at my books. Although I can't find the bit that said about swapping to high,low but I'm sure that I did read it somewhere. It does say that the device only gives the bits 1-7 and bit 0 is always 0 giving 128 possible addresses. Dave - Original Message - From: David Brant [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 10:49 PM Subject: Re: Short, short questions This was based on info from a book called z-80 Workshop manual by E.A Parr. The I register gives the high part of the table and the hardware gives the low part to the table then takes that word for the service routine. So if you start from one byte before the table and use the same address for all entries and over run it by one it will work. My demo of a full scrolling football pitch used this system, which I believe you saw many years a go. Dave - Original Message - From: Andrew Collier [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 9:50 PM Subject: Re: Short, short questions Hi, I'm sceptical about this claim. I've never heard anybody say that the vector formed is big-endian - it's just you don't know the byte offset from which the interrupt vector will be fetched. (As Edwin says, it is usually 255 - which is odd so your 1-aligned table will usually work - but I don't know that Sam's hardware guarantees this). So the high byte comes from I, the low byte from the data bus; this forms a 16 bit address which will be incremented once (which is why the table needs 257 bytes, not 256). You could, at least in theory, read the vector address from even or odd overlapping entries, which is why the usual strategy is to pick a vector address whose low and high bytes are the same. The last IM2 interrupt routine I wrote looked something like this: ds ALIGN 256 IM2TABLE: equ $ IM2BYTE: equ im2table/256 IM2TARGETBYTE: equ IM2BYTE+1 for 257, DB IM2TARGETBYTE IM2TARGET: equ 257*IM2TARGETBYTE ds IM2TARGET-$ EX AF,AF' ... Andrew On 20 May 2008, at 21:16, David Brant wrote: Mode 2 uses a table with 128 word address but as byte high,byte low not the normal low, high bytes So if you set your org/dump address to ??FF (i.e. ??00-1) and then do DEFWmode2.i,mode2.i so you have 129 words. mode2.i: di pushaf ina,(status.int) . . ei ret - Original Message - From: Andrew Collier [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 3:22 PM Subject: Re: Short, short questions The usual strategies are to use mode 1, or to use mode 2 with a 257- byte table all containing the same byte. -- --- Andrew Collier http://www.intensity.org.uk/ --- -- -- --- Andrew Collier http://www.intensity.org.uk/ --- --
Re: Short, short questions
All 8 bits are used for LSB of the vector. The part where bit 0 always is zero is when one of the Z80's IO chips is connected (PIO,SIO,CTC etc) is connected. Edwin - Original Message - From: David Brant [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Wednesday, May 21, 2008 1:02 AM Subject: Re: Short, short questions I've just been looking at my books. Although I can't find the bit that said about swapping to high,low but I'm sure that I did read it somewhere. It does say that the device only gives the bits 1-7 and bit 0 is always 0 giving 128 possible addresses. Dave - Original Message - From: David Brant [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 10:49 PM Subject: Re: Short, short questions This was based on info from a book called z-80 Workshop manual by E.A Parr. The I register gives the high part of the table and the hardware gives the low part to the table then takes that word for the service routine. So if you start from one byte before the table and use the same address for all entries and over run it by one it will work. My demo of a full scrolling football pitch used this system, which I believe you saw many years a go. Dave - Original Message - From: Andrew Collier [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 9:50 PM Subject: Re: Short, short questions Hi, I'm sceptical about this claim. I've never heard anybody say that the vector formed is big-endian - it's just you don't know the byte offset from which the interrupt vector will be fetched. (As Edwin says, it is usually 255 - which is odd so your 1-aligned table will usually work - but I don't know that Sam's hardware guarantees this). So the high byte comes from I, the low byte from the data bus; this forms a 16 bit address which will be incremented once (which is why the table needs 257 bytes, not 256). You could, at least in theory, read the vector address from even or odd overlapping entries, which is why the usual strategy is to pick a vector address whose low and high bytes are the same. The last IM2 interrupt routine I wrote looked something like this: ds ALIGN 256 IM2TABLE: equ $ IM2BYTE: equ im2table/256 IM2TARGETBYTE: equ IM2BYTE+1 for 257, DB IM2TARGETBYTE IM2TARGET: equ 257*IM2TARGETBYTE ds IM2TARGET-$ EX AF,AF' ... Andrew On 20 May 2008, at 21:16, David Brant wrote: Mode 2 uses a table with 128 word address but as byte high,byte low not the normal low, high bytes So if you set your org/dump address to ??FF (i.e. ??00-1) and then do DEFWmode2.i,mode2.i so you have 129 words. mode2.i: di pushaf ina,(status.int) . . ei ret - Original Message - From: Andrew Collier [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 3:22 PM Subject: Re: Short, short questions The usual strategies are to use mode 1, or to use mode 2 with a 257- byte table all containing the same byte. -- --- Andrew Collier http://www.intensity.org.uk/ --- --
Re: Short, short questions
I thought the idea of mode2 was you could have different vectors for different devices connected well this throws a spanner in the works. But then again is there any hardware for the SAM that uses them? I think it must have been an old spectrum book that said this about swapping high,low bytes. After a little test and using old brain this is wrong. Dave - Original Message - From: Edwin Blink [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Wednesday, May 21, 2008 5:34 AM Subject: Re: Short, short questions All 8 bits are used for LSB of the vector. The part where bit 0 always is zero is when one of the Z80's IO chips is connected (PIO,SIO,CTC etc) is connected. Edwin - Original Message - From: David Brant [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Wednesday, May 21, 2008 1:02 AM Subject: Re: Short, short questions I've just been looking at my books. Although I can't find the bit that said about swapping to high,low but I'm sure that I did read it somewhere. It does say that the device only gives the bits 1-7 and bit 0 is always 0 giving 128 possible addresses. Dave - Original Message - From: David Brant [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 10:49 PM Subject: Re: Short, short questions This was based on info from a book called z-80 Workshop manual by E.A Parr. The I register gives the high part of the table and the hardware gives the low part to the table then takes that word for the service routine. So if you start from one byte before the table and use the same address for all entries and over run it by one it will work. My demo of a full scrolling football pitch used this system, which I believe you saw many years a go. Dave - Original Message - From: Andrew Collier [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 9:50 PM Subject: Re: Short, short questions Hi, I'm sceptical about this claim. I've never heard anybody say that the vector formed is big-endian - it's just you don't know the byte offset from which the interrupt vector will be fetched. (As Edwin says, it is usually 255 - which is odd so your 1-aligned table will usually work - but I don't know that Sam's hardware guarantees this). So the high byte comes from I, the low byte from the data bus; this forms a 16 bit address which will be incremented once (which is why the table needs 257 bytes, not 256). You could, at least in theory, read the vector address from even or odd overlapping entries, which is why the usual strategy is to pick a vector address whose low and high bytes are the same. The last IM2 interrupt routine I wrote looked something like this: ds ALIGN 256 IM2TABLE: equ $ IM2BYTE: equ im2table/256 IM2TARGETBYTE: equ IM2BYTE+1 for 257, DB IM2TARGETBYTE IM2TARGET: equ 257*IM2TARGETBYTE ds IM2TARGET-$ EX AF,AF' ... Andrew On 20 May 2008, at 21:16, David Brant wrote: Mode 2 uses a table with 128 word address but as byte high,byte low not the normal low, high bytes So if you set your org/dump address to ??FF (i.e. ??00-1) and then do DEFWmode2.i,mode2.i so you have 129 words. mode2.i: di pushaf ina,(status.int) . . ei ret - Original Message - From: Andrew Collier [EMAIL PROTECTED] To: sam-users@nvg.ntnu.no Sent: Tuesday, May 20, 2008 3:22 PM Subject: Re: Short, short questions The usual strategies are to use mode 1, or to use mode 2 with a 257- byte table all containing the same byte. -- --- Andrew Collier http://www.intensity.org.uk/ --- --
