I have a count in a double register.No fast way.
Is there a way to decrement and test for 0 without using the A register?
NOTE! 8-bit increment/decrement does NOT affect the carry. Only sign or zero may be detected after those. If you use zero as the trigger for a dual 8-bit decrement loop (i.e. accomplishing a 16-bit decrement using two 8-bit decrements) you need to takeinto account that zero is not the same as carry and you will have to adjust your starting counts accordingly.
Douglas Miller <durgadas311@gmail.com> wrote:take into account that zero is not the same as carry and you will have to adjust your starting counts accordingly.
NOTE! 8-bit increment/decrement does NOT affect the carry. Only sign or zero may be detected after those. If you use zero as the trigger for a dual 8-bit decrement loop (i.e. accomplishing a 16-bit decrement using two 8-bit decrements) you need to
You are right. My mistake. Using only the detection of zero a possible solution is doing a 16 bit decrement and for the low and high bytes
an increment followed by a decrement. So the value of the register does
not change but only the flags are set. A test costs 36 cycles if
both registers have to be checked or at least 18 cycles if the low byte
is not zero after the 16 Bit decrement.
You are right. My mistake. Using only the detection of zero a possible solution is doing a 16 bit decrement and for the low and high bytes
an increment followed by a decrement. So the value of the register does
not change but only the flags are set. A test costs 36 cycles if
both registers have to be checked or at least 18 cycles if the low byte
is not zero after the 16 Bit decrement.
--
Dipl.-Inform(FH) Peter Heitzer, peter....@rz.uni-regensburg.de
On 2/08/2023 9:48 pm, Peter Heitzer wrote:take into account that zero is not the same as carry and you will have to adjust your starting counts accordingly.
Douglas Miller <durgadas311@gmail.com> wrote:
NOTE! 8-bit increment/decrement does NOT affect the carry. Only sign or zero may be detected after those. If you use zero as the trigger for a dual 8-bit decrement loop (i.e. accomplishing a 16-bit decrement using two 8-bit decrements) you need to
You are right. My mistake. Using only the detection of zero a possible
solution is doing a 16 bit decrement and for the low and high bytes
an increment followed by a decrement. So the value of the register does
not change but only the flags are set. A test costs 36 cycles if
both registers have to be checked or at least 18 cycles if the low byte
is not zero after the 16 Bit decrement.
My initial thought was to save/restore A e.g.
push h
mov l,a
mov a,e
ora d
mov a,l
pop h
but hoping there was something smarter. I later discovered a logical error >in my code and after correcting it the problem went away...
My initial thought was to save/restore A e.g.
push h
mov l,a
mov a,e
ora d
mov a,l
pop h
but hoping there was something smarter. I later discovered a logical error in my code and after correcting it the problem went away...
dxforth <dxforth@gmail.com> wrote:
[Re: testing if a register pair is zero on an 8080]
My initial thought was to save/restore A e.g.
push h
mov l,a
mov a,e
ora d
mov a,l
pop h
but hoping there was something smarter. I later discovered a logical error >> in my code and after correcting it the problem went away...
That makes it sound like you don't need the test any more, but I may
as well still say that self-modifying code would be slightly faster:
sta aop+1
mov a,e
ora d
aop: mvi a,0
8080 or 8085? If 8085, there is the undocumented instruction "DSUB" (08h), where:In 8080:
HL = HL - BC (Z, S, P, CY, AC and X5, V all flag receives influence)
LXI H some number
LXI B 1
DSUB
JZ (or JNZ) somewhere
https://electronicerror.blogspot.com/2007/08/undocumented-flags-and-instructions.html
Mark Lougheed <mdlou...@gmail.com> wrote:Depending on usage, one option is to pre increment the two bytes of the count as in
8080 or 8085? If 8085, there is the undocumented instruction "DSUB" (08h), where:
HL = HL - BC (Z, S, P, CY, AC and X5, V all flag receives influence)
LXI H some number
LXI B 1
DSUB
JZ (or JNZ) somewhere
https://electronicerror.blogspot.com/2007/08/undocumented-flags-and-instructions.htmlIn 8080:
PUSH B
LXI B 0ffffh
DAD B
POP B
JC iszero
Needs 51 cycles and the counter to be tested must be in HL.
--
Dipl.-Inform(FH) Peter Heitzer, peter....@rz.uni-regensburg.de
I have a count in a double register.
Is there a way to decrement and test for 0 without using the A register?
On Wednesday, 2 August 2023, dxforth wrote:
I have a count in a double register.
Is there a way to decrement and test for 0 without using the A register?
If the target is 8085 then there is an undocumented flag and jump instruction available.
The flag is usually called K and I often use the JP NK instruction to emulate a z80 LDIR instruction. The K flag is set on 0 to 0xFFFF so the loop counter needs to be pre-decremented.
Usage xamples are here. https://github.com/RC2014Z80/RC2014/blob/master/ROMs/CPM-IDE/acia85cf/cpm22preamble.asm#L34
Of course this is only going to work if your target is actually 8085.
On 8/08/2023 2:11 pm, Phillip Stevens wrote:
Of course this is only going to work if your target is actually 8085.
A pity Intel didn't seriously look at an extended instruction set.
Why Faggin left?
On Tuesday, 8 August 2023 at 22:54:05 UTC+10, dxforth wrote:really accelerated math (given no hardware multiply).
On 8/08/2023 2:11 pm, Phillip Stevens wrote:
Of course this is only going to work if your target is actually 8085.
A pity Intel didn't seriously look at an extended instruction set.
A bit OT but since you asked. The enhancements Intel made to the 8085, and then decided not to support, made the 8085 almost perfect imho. Undocumented stack access using DE instructions is much faster than IX/IY in the z80, and the 16 bit rotations
Using these enhancements, and a “native” C compiler, the 8085 actually beats the z80 in some of our z88dk benchmarks.
So one excellent code table (256) of instructions is a pretty nice design. https://feilipu.me/2021/09/27/8085-software/
Why Faggin left?
Wasn’t there, but I guess it is easier to get rich working for yourself. 😊
Something he perhaps wouldn’t have achieved at Intel.
Using these enhancements, and a “native” C compiler, the 8085 actually beats the z80 in some of our z88dk benchmarks.
So one excellent code table of (256) instructions is a pretty nice design.I assumed any undoc instructions would be either side-effects or planned but broken in some way. From what you say this doesn't appear to be the case here.
https://feilipu.me/2021/09/27/8085-software/
Sounds like it was an executive decision to leave them out. Unfortunately the
result is the same - software that uses undoc instructions have a certain odour
to them.
Using these enhancements, and a “native” C compiler, the 8085 actually beats the z80 in some of our z88dk benchmarks.I assumed any undoc instructions would be either side-effects or planned but >> broken in some way. From what you say this doesn't appear to be the case here.
So one excellent code table of (256) instructions is a pretty nice design. >>> https://feilipu.me/2021/09/27/8085-software/
Sounds like it was an executive decision to leave them out. Unfortunately the
result is the same - software that uses undoc instructions have a certain odour
to them.
I wouldn’t worry too much. Tundra Semiconductor licensed the 80c85 design and published them as “enhanced instructions” in their datasheet. Complete with their own mistakes and mis interpretation.
http://images.100y.com.tw/pdf_file/34-TUNDRA-CA80C85B.pdf
Ken Shirriff covers the reverse engineering in some details and corrects the error in the Tundra datasheet.
His whole series is great reading on the 8085.
http://www.righto.com/2013/02/looking-at-silicon-to-understanding.html?m=1
So safe to use with no hesitation or odour. :-)
If you're willing to alter the count you can do the 16 bit loop as two 8 bit tests which will be faster. For example, suppose your 16 bit loop count is in DE. This code will execute the loop DE times:iterations and then after that it will do 256 iterations D times (the original value of D). Check it with $0003 and $0103 to get the idea.
; transform DE to split count
dcx d
inr d
inr e
loop:
; ... processing here
dcr e
jnz loop
dcr d
jnz loop
To see how it works consider the two cases going in where E is zero and not zero. Or, equivalently, when DE is a exact multiple of 256 or not. Note that the value of E never changes. If E is not zero then 1 is added to D. The inner loop will do E
If E is zero then D is not altered either. The inner loop will always be 256 iterations and will be executed D times. Checking $0100 and $0000 will show the correctness. Like with a conventional 16 bit loop, $0000 means 65536 iterations.
If you're willing to alter the count you can do the 16 bit loop as two 8 bit tests which will be faster. For example, suppose your 16 bit loop count is in DE. This code will execute the loop DE times:iterations and then after that it will do 256 iterations D times (the original value of D). Check it with $0003 and $0103 to get the idea.
; transform DE to split count
dcx d
inr d
inr e
loop:
; ... processing here
dcr e
jnz loop
dcr d
jnz loop
To see how it works consider the two cases going in where E is zero and not zero. Or, equivalently, when DE is a exact multiple of 256 or not. Note that the value of E never changes. If E is not zero then 1 is added to D. The inner loop will do E
If E is zero then D is not altered either. The inner loop will always be 256 iterations and will be executed D times. Checking $0100 and $0000 will show the correctness. Like with a conventional 16 bit loop, $0000 means 65536 iterations.If a count of zero is valid, then the tests need to be moved to the start of the loop
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