moving arrays -- 1 completion of ideas needed for a general and complete program final concepts...
TRANSCRIPT
Moving Arrays -- 1
Completion of ideas needed for a general and complete program
Final concepts needed for FinalReview for Final – Loop efficiency
04/20/23DMA , Copyright M. Smith, ECE, University of Calgary, Canada
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Tackled today
Declaring and initializing arrays off the stack – Review and a little bit of new Useful for background DMA tasks Useful for minimizing total memory used in
non-general program Declaring arrays and variables on the stack
– Review and a little bit of new Re-entrant code and thread safe
Demonstrating memory to memory DMA
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Declaring fixed arrays in memory – not on the stack
short foo_startarray[40];short far_finalarray[40];
void HalfWaveRectifyASM( ) {// Take the signal from foo_startarray[ ] and rectify the signal// Half wave rectify – if > 0 keep the same; if < 0 make zero// Full wave rectify – if > 0 keep the same; if < 0 then abs value// Rectify startarray[ ] and place result in finalarray[ ]
for (int count = 0; count < 40; count++) {if (foo_startarray[count] < 0) far_finalarray[count] = 0;else far_finalarray[count] = foo_startarray[count];
}}
The program code is the same – but the data part is not
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First attemptto get correctanswer
.section data1
Tells linker to place this stuff in memory map location data1
.align 4 – adjust address to end in 0, 4, 8 or CWe know processor works best when we start things on a boundary between groups of 4 bytes
[N * 2] We need N short ints
We know the processor works with address working in bytes. Therefore need N * 2 bytes sounds sensible
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“wrong approach” – does not match with what C / C++ does with memory
20 bytes (16 bits) for
N short value in C++ = N * 2 bytes
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“Correct approach was NOT what I expected”
ASM Array with space for N long ints .var arrayASM[N]; better .byte4 arrayASM[N];
ASM Array with space for N short ints var arrayASM[N / 2]; better .byte2 arrayASM[N};
ASM Array with space for N chars var arrayASM[N / 4]; better .byte arrayASM[N];
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Better answer is “Look at the assembler manual”
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Improving what we did before
Big warning – external array initialization occurs on “reload” of your program code and NOT on “restart” of your program code (WHY?) Understanding why this is true and why it is a problem will solve many issues when programming
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04/20/23DMA , Copyright M. Smith, ECE, University of Calgary, Canada
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When DMA might be useful-- Video manipulation
ProgramWait for picture 1 to come in – video-inProcess picture 1 – lots of
mathematics perhapsWait for picture 1 to be transmitted –
video out Spending a lot of time waiting rather
than doing
04/20/23DMA , Copyright M. Smith, ECE, University of Calgary, Canada
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When DMA might be useful-- Double Buffering
Program1. Wait for picture 2 memory to fill – video-in2. Picture 3 comes into memory – background DMA task from input
Process picture 2 – place result into picture 0 location3. Picture 4 comes into memory – background DMA task from input
Process picture 3 – place result into picture 1 locationTransmit picture 0 – background DMA task to output
4. Picture 0 comes into memory – background DMA task from inputProcess picture 4 – place result into picture 2 locationTransmit picture 1– background DMA task to output
5. Picture 1 comes into memory – background DMA task from inputProcess picture 0 – place result into picture 3 locationTransmit picture 2 – background DMA task to output
6. Picture 2 comes into memory – background DMA task from inputProcess picture 1 – place result into picture 4 locationTransmit picture 3– background DMA task to output
7. REPEAT STEPS FOR EVER
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We are only going to look at a simple DMA task
Normal code when trying to move data from one location to another Number of simple examples in Lab. 3 using SPI interface1) P0 address of start_array[0];2) P1 address of final_array[0];3) R0 number of data items to be transferred needed to transfer4) R1 How many values already transferred
5) R1 = 0;LOOP: 6) CC = R0 <= R17) IF CC JUMP DONE:8) R2 = [P0++]; VERY BIG PIPELINE9) [P1++] = R2; LATENCY ISSUES10) JUMP LOOP; MANY INTERNAL
PROCESSOR STALLS ON DATA BUSDONE: WHILE WAIT FOR R2 TO BE
Must wait to Do something else READ, STORED and then TRANSMITTED
INSTRUCTION BUS STALLS EVERY TIMETHE CODEJUMPS-- LOSE 4 CYCLES
04/20/23DMA , Copyright M. Smith, ECE, University of Calgary, Canada
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We are only going to look at a simple DMA task
DMA special hardware that works without the processor 1) DMA_source_address_register address of start_array[0]; 2) DMA_destination_address_register address of final_array[0];3) DMA_max_count_register max-value needed to transfer 4) DMA_count_register How many values already transferred
R1 = 0;LOOP:
CC = R0 <= R1IF CC JUMP DONE: 5) DMA_enable = true
R2 = [P0++]; DMA transfer happen in background [P1++] = R2; Miminized pipeline issuesJUMP LOOP;
DONE:Do something else Processor can do something else
immediately while DMA hardware handles all the memory transfers WITHOUT PROCESSOR HELP.
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Write some tests so we know how to proceed -- Test 1
Is DMA useful when the arrays being moved are in the processor’sinternal memory and placed on the stack as with this code
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Write some test so we know how to proceed -- Test 2
IS DMA useful when both the arrays are placed in external memorySDRAM is needed for large video images
SDRAM -- MANY MEGS AVAILABLE
SDRAM addresses hard-coded in this example
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Write some test so we know how to proceed -- Test 3
Most probable way to use DMA – Store video arrays in SLOW external memoryMove to FAST internal memory for processing, put result back into external
SDRAM addresses hard-coded in this example
WAIL -- Can use compilersection (“SDRAM”) syntax
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Some resultsCode details later
CompilerDebug Mode
CompilerRelease Mode
L1 L1Internal memory
8748 625
L1 L1 DMA 6579 6477DMA slower
SDRAM SDRAMexternal
39132 28200
SDRAM SDRAM DMA
12175 12090
SDRAM L1 DMA 5265 4836
SDRAM L1 DMAL1 SDRAM DMA
9792 9276
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Memory to memory move Debug Code
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Review for final
A) What happened here?
B) What happened here?
C) What happened here?
E) What happened here?
F) Determine loop efficiency in terms of instructions in terms of cycles / read_write op
D) Why did this happen?
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Answer questions
A B C D E
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Review for finalInternal memory to Internal memory
F) Determine loop efficiency in terms of cycles / read_write op
internal memory -> internal memory
size was 300 Useful reads 300 Useful writes 300
Cycles 8748 as measured
8748 / 600 = 14.58
Why not an exact number?Instructions in loop? 19Total # of reads / write 9 / loop2700 read / writes – around 3 cycles
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Review for finalSDRAM to SDRAM
F) Determine loop efficiency in terms of cycles / read_write op
SDRAM external -> SDRAM memory Useful reads / writes 300 each
Cycles 39132 as measured 39132 / 600 = 65.22
Why not an exact number?Instructions in loop? 19Total # of reads / write 9 / loop7 * 300 read / writes internal2 * 300 read / writes external
Time r/w external = 39132 – 2100*3 33000 / 600 = 5.5 cycles Factor of 2 slower
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Memory to memory moveRelease Mode
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Review for final
A) What happened here?
B) What happened here?
C) What happened here?
E) What happened here?
F) Determine loop efficiency in terms of instructions in terms of cycles / read_write op
D) Why did this happen inside loop?
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Answer questions
A B C D E
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F) Determine loop efficiency in terms of cycles / read_write op
internal memory -> internal memory
size was 300 Useful reads 300 Useful writes 300
Cycles 625 as measured
625 / 600 = 1.05
Why not an exact number?Instructions in loop? 4300 * 4 = 1200
WE WOULD EXPECT 1200 cycles!!!!Where did the difference go?
Release modeinternal to internal
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F) Determine loop efficiency in terms of cycles / read_write op
SDRAM -> internal memory
size was 300 Useful reads 300 Useful writes 300
Cycles 28200 as measured
28200 / 600 = 47
SDRAM access 47 cyclesL1 memory 1 cycle
Would make sense to process in L1 memory – so move SDRAM to L1 to process
Release modeexternal to external
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F) Determine loop efficiency in terms of cycles / read_write op
SDRAM -> internal memory
size was 300 Useful reads 300 Useful writes 300
Cycles 4836 as measured 300 of those are L1 writes Leaving 4500
4500 / 300 = 15
SDRAM read before 47 cyclesSDRAM read now 15 cycles L1 -> L1 1 cycle
Would make sense to process in L1 memory – so move SDRAM to L1 to process
Loads of overhead in SDRAM to SDRAM
External to internal
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Tackled today
Review of handling external arrays (global arrays) from assembly code Arrays declared in another file Arrays declared in this file -- NEW Needed for arrays used by ISRs
Arrays declared on the stack Pointers passed as parameters to a subroutine Can’t use arrays on the stack when used by ISR
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Information taken from Analog Devices On-line Manuals with permission http://www.analog.com/processors/resources/technicalLibrary/manuals/
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