1. Memory Mapped Systems 2. Adding Unsigned Numbers

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1 1 Memory Mapped Systems 2 Adding Unsigned Numbers 1

2 1 Memory Mapped Systems Our system uses a memory space Address bus is 16-bit locations Data bus is 8-bit 2 Adding Unsigned Numbers 2

Memory: RAM, ROM, FLASH, etc Input/Output

3 Our system uses a memory space Address bus is 16-bit locations Data bus is 8-bit Controller Memory Space: locations x 8-bit Central Processing Unit (CPU, ie, μp) Memory: RAM, ROM, FLASH, etc Input/Output (I/O) Controllers ABUS CBUS DBUS 3

4 WHAT IS A MEMORY MAPPED SYSTEM? Each device which the μp controls and uses is assigned a distinct address range in the memory space A device is a memory chip or an I/O controller (with a bunch of registers) The assignment is called mapping a device to the memory space Once a device is mapped into the memory space, when you write a program you will know where the device is located, and therefore, you will know what addresses to use in instructions to read data from and write data to the device For example, the figure to the right shows the address range of each device, ie, shows where each device is mapped in the memory space Note that some areas in the memory space may not be used (as indicated by NOT USED ) FF FF F 0610 EFFF F000 FFFF RAM1 RAM2 PORT IO NOT USED ROM 4

5 Example (Continued) This view shows the devices on the system bus, but does not show the address ranges of the chips Central Processing Unit (CPU, ie, μp) 512x8 RAM1 1Kx8 RAM2 16x8 I/O Controller 4Kx8 ROM 16 8 ABUS CBUS DBUS SYSTEM BUS 5

6 A MEMORY MAPPED SYSTEM EXAMPLE 2 6

7 7 MEMORY MAPPED SYSTEMS OTHER EXAMPLES USER RAM PORT IO FLASH ROM INTERRUPT VECCTORS FF NOT USED FF A000 NOT USED 0804 BFFF DFFF E000 FFF7 FFF8 FFFF USER RAM ROM INTERRUPT VECCTORS FFF NOT USED 8000 E000 FFF7 FFF8 FFFF PORT IO DFFC DFFF RAM1 PORT IO FLASH INTERRUPT VECCTORS FF RAM FF F NOT USED 0810 DFFF E000 FFF7 FFF8 FFFF

LAB SYSTEM (PARTIAL) MEMORY MAP FLASHStart EQU $4000 ;Address to place my code/constant data FLASHEnd EQU $7FFF RAMStart EQU $1400 ;Address to place my variables and stack RAMEnd EQU $3FFF ORG

8 LAB SYSTEM (PARTIAL) MEMORY MAP FLASHStart EQU $4000 ;Address to place my code/constant data FLASHEnd EQU $7FFF RAMStart EQU $1400 ;Address to place my variables and stack RAMEnd EQU $3FFF ORG RAMStart ;Allocate space for my variables num1 DCB $FF ;Install 1st number at memory location $1400 num2 DCB $03 ;Install 2nd number at memory location $1401 MSresult DSB 1 ;Allocate space to store MS Byte of result LSresult DSB 1 ;Allocate space to store LS Byte of result ORG RAMStart ;Allocate space for my stack ;I have no stack for this program ORG FLASHStart ;Put my program code into FLASH memory, ;starting at $4000 Entry CLRA LDAB num1 ADDB num2 BCC skip LDAA #1 skip STAA MSresult STAB LSresult here BRA here ;Stop the program by looping continually here ORG $FFFE ;Address of Reset Vector DCW Entry ;Install my Reset Vector 8

9 SIZE AND NUMBER OF CHIPS Amount of RAM, ROM, FLASH, and other memories depend on the application A home security system may need a small amount of RAM, ROM, and a few locations for IO controllers A sound recording system and data logging application may need large amount of RAM, small ROM (program), and a few locations for IO controllers You will find that the course projects for this course each need a different amount and type of memory chips 9

SYSTEM BUS ARCHITECTURE von Neumann architecture μp

ROM 8 8 8 8 8 Single bus for both program and data

Program and data must be accessed at different times

to implement I/O 1 8 ABUS CBUS I/O 2 RAM1 RAM2 8 8 8

architecture Separate program memory-bus and data

separate buses DBUS DATA BUS PROGRAM BUS DBUS Program

10 SYSTEM BUS ARCHITECTURE von Neumann architecture μp Bus Master 16 8 ABUS CBUS DBUS RAM1 RAM2 I/O 1 I/O 2 ROM Single bus for both program and data May have different or same chips for program and data Program and data must be accessed at different times This course uses von Neumann SYSTEM BUS Less complex to implement I/O 1 8 ABUS CBUS I/O 2 RAM1 RAM μp Bus Master 16 8 ROM 8 ABUS CBUS Harvard architecture Separate program memory-bus and data memory-bus Program and data are accessed from separate buses DBUS DATA BUS PROGRAM BUS DBUS Program and data may be accessed at the same time More complex to implement 10

11 WHERE ARE PROGRAM INSTRUCTIONS STORED? A ROM chip is generally used to store the initial program (aka reset routine) to be run on the microprocessor Examples: EEPROM, FLASH, PROM Program needs to be preserved when power is shut off When developing microprocessing applications, program instructions may be stored in RAM to facilitate development, debugging, and testing Once fully developed, debugged, and tested, the program will be burnt into a ROM chip, located in the memory map 11

12 1 Memory Mapped Systems 2 Adding Unsigned Numbers 12

ADDING UNSIGNED NUMBERS When you add two 8-bit unsigned numbers, you need to consider the most significant carry out: For example, if you do: $FF + $01 = 255 + 1, the 8-bit result would be $00, and

13 ADDING UNSIGNED NUMBERS When you add two 8-bit unsigned numbers, you need to consider the most significant carry out: For example, if you do: $FF + $01 = , the 8-bit result would be $00, and the C=1 in the CCR Therefore, your answer should be $0100 = 256 LDAB $0800 ADDB $0801 BCS Prepend$01 LDAA #$00 BRA SKIP Prepend$01 LDAA #$01 SKIP STAA $0802 STAB $0803 DONE BRA DONE CLRA LDAB $0800 ADDB $0801 ADCA #$00 STAA $0802 STAB $0803 DONE BRA DONE Two different programs that do the same thing Next: how to add a whole bunch of numbers? 13

14 USER RAM Given USER RAM from 0100 to 01FF (256 bytes; how did I get 256?) Constraint: program and data are to be stored in the same memory chip: USER RAM 0100 USER RAM CHIP Program Code Area Need to keep program code and data separate Why? (See next slide for example of what can go wrong) Memory allocation plan: Program code to start at 0100 Note: size of program will not be known until it has been completed Data Area Place data at other end of the USER RAM chip, as far as possible from the program, to allow for maximum size of program area and maximum size of data area 01FF 14

15 Aside: why must program code and data be separated? Ex of what can go wrong Line Address Machine Code Assembly Language Comments 1 00FE B LDAA $0107 Load contents of $0107 into Accumulator A (A=7E) BB ADDA $0108 Add the contents of $0108 to Accumulator A (A=7F) B STAA $0109 Stores the result at $0109 ($0109) = 7F E Supposed to be start of data area (variable 1) Variable Result There are two major problems here: 1 A program end was not defined 2 Data was placed in the program code area Developer mistakenly places the data 7E and 01 into locations $ 0107 and $0108, respectively, and uses $0109 to store the result Without a program end being defined, these memory locations are part of the program After executing the STAA $0109 instruction, the microprocessor fetches the next instruction, expecting an opcode, and puts the value fetched $7E in the OCR It (ie, $7E) is interpreted as the JMP EXT instruction, which is executed as such To execute this instruction, the microprocessor forms the address of where to jump to by copying the next two bytes ($01 $7F) into the THR Finally, the microprocessor jumps to location $017F, which contains undefined program instructions and the program crashes 15

16 WHAT IS THE MAXIMUM POSSIBLE SUM? 46 bytes to be added together should be stored in USER RAM USER RAM 0100 To determine the maximum possible sum, assume all data bytes are maximum value, ie, 255; then, the maximum unsigned sum would be: 46 * 255 = 11,730 Program Code Area The sum must be stored in two locations (2 Bytes are needed to represent the sum) Note: 2 Bytes can count up to 65,535 (FFFF) 01FD 01FE 01FF RESULT HB RESULT LB Data Area Let s plan to store the result at 01FE:01FF 16

17 WHAT SHOULD BE THE STARTING ADDRESS OF THE DATA? Ending address (EA) is $01FD: solve for starting address (SA): 0100 EA = SA (Why 1?) Program Code Area SA = EA 45 = 01FD 002D = 01D0 Starting Address (SA) SA = 01D0 SA = 2D 16 SUBTRACT A B C D E F Ending Address (EA) 01FD 01FE 01FF RESULT HB RESULT LB 46 numbers to be Added Result 4-bit number wheel unrolled 17

18 Now, populate the memory with 46 bytes to be added Now we have a List (Array, Block) of data to add 01D0 USE RAM CHIP 01 FF 7F AB 81 Note that data in memory may be interpreted differently: Example: AB = 171 or -85 We interpret the numbers as unsigned for this example 01FD 01FE 01FF 3F RESULT HB RESULT LB Data Area 18

19 MAIN IDEA OF HOW TO ADD THE LIST OF NUMBERS ACCUMULATOR B bit Binary Number Wheel ACCUMULATOR A bit Binary Number Wheel D0 USER RAM CHIP 01 FF 7F AB COUNTS THE CARRY OUTs OF ACCUMULATOR A ADDS ALL BYTES OF DATA IN THE LIST MOST SIGNIFICANT PART OF SUM ACCB:ACCA LEAST SIGNIFICANT PART OF SUM 01FD 01FE 01FF 3F RESULT HB RESULT LB Ex: 01 + FF = 00 with C=1 B = 01 and A = 00 ACCB:ACCA =

20 INITIAL SOLUTION ATTEMPT POOR SOLUTION: WON T WORK 01D0 USER RAM CHIP 01 FF 7F AB 81 Clear high byte of result (ACCB = 0) Load ACCA with the 1 st number: ACCA (01D0) Add the next number: ACCA ACCA + (01D1) Add the carry to ACCB: ACCB $00 + C CLRB LDAA $01D0 ADDA $01D1 ADCB #$00 ADDA $01D2 Repeat preceding two instructions 44 more times, with appropriate changes in the address (01XX) Add the 46 th number: ACCA ACCA + (01FD) ADCB #$00 ADDA $01FD Add the carry to ACCB: ACCB $00 + C ADCB #$00 01FD 01FE 01FF 3F RESULT HB RESULT LB Store the high byte of result: (01FE) ACCB Store the low byte of result: (01FF) ACCA STAB STAA $01FE $01FF Done DONE BRA DONE 20

21 PROGRAM LISTING FILE *LST Line Address Machine Code Assembly Language Comments F CLRB Make high byte of result = zero B6 01 D0 LDAA $01D0 Load contents of $01D0 into Accumulator A BB 01 D1 ADDA $01D1 Add the contents of $01D1 to Accumulator A C9 00 ADCB #$00 Add 00 and the carry to Accumulator B BB 01 D2 ADDA $01D2 Add the contents of $01D2 to Accumulator A 6 010C C9 00 ADCB #$00 Add 00 and the carry to Accumulator B 91 01E0 BB 01 FD ADDA $01FD Add the contents of $01FD to Accumulator A 92 01E3 C9 00 ADCB #$00 Add 00 and the carry to Accumulator B 93 01E5 F7 01 FE STAB $01FE Store HB of the result in location $01FE 94 01E8 B7 01 FF STAA $01FF Store LB of the result in location $01FF 95 01EB 20 FE DONE BRA DONE Infinite loop here Note: Called a listing file because it additionally lists the contents of memory for the program 21

22 ASIDE: DIFFERENT WAYS OF VIEWING MEMORY For Readability: Since each line represents one instruction Address Machine Code F 0101 B6 01 D BB 01 D C BB 01 D2 010C 01E0 BB 01 FD 01E3 C E5 F7 01 FE 01E8 B7 01 FF 01EB 20 FE BYTE Organized Address Machine Code F 0101 B D BB D C BB 010A 01 D2 01EB 20 01EC FE Actual: Each memory location holds 8 voltages (either 5 V and 0 V), but we represent the voltages by binary symbols Address Machine Code A EB EC

23 PROGRAM Address Machine Code F 0101 B6 01 D BB 01 D C BB 01 D2 010C C E0 BB 01 FD 01E3 C E5 F7 01 FE 01E8 B7 01 FF 01EB 20 FE PROGRAM SIZE 01EC = 00ED = 237 Bytes 237 Locations in Memory PROGRAM CYCLES 1x2 1x4 45x4 45x2 2x4 1x3 Total: 287 For 2 MHz Clock: 1435 μs 23

24 WHY SOLUTION 1 WON T WORK: Address Machine Code F 0101 B D BB D C BB 010A 01 D2 01EB 20 01EC FE Starting Address (SA) Ending Address (EA) D0 01EC 01FD 01FE 01FF RESULT HB RESULT LB Program Code Area Data Area NOT ENOUGH ROOM FOR PROGRAM IN GIVEN MEMORY CHIP! 24

25 SOLUTION 2 LOOP N Start A=0 B=0 X = $01D0 A=A+(X) B=B + Carry X=X+1 X==$01FE? Y X 01D0 USER RAM CHIP 01 FF 7F AB 81 CLRA CLRB LDX #$01D0 Loop ADDA $00,X ADCB #$00 INX CPX BNE STAB #$01FE Loop $01FE Store high byte of result (ie, B) at location $01FE Store low byte of result (ie, B) at location $01FF 01FD 01FE 01FF 3F RESULT HB RESULT LB STAA $01FF HERE BRA HERE End 25

26 PROGRAM Address Machine Code F F 0102 CE 01 D AB C A 8C 01 FE 010D 26 F6 010F F7 01 FE 0112 B7 01 FF FE PROGRAM SIZE = 0016 = 22 Bytes 22 Locations in Memory CLRA CLRB LDX #$01D0 Loop ADDA 00,X ADCB #$00 INX CPX #$01FE BNE Loop STAB $01FE STAA $01FF HERE BRA HERE PROGRAM CYCLES 2x2 1x3 46x4 46x2 46x2 46x4 46x3 2x4 1x3 Total: 708 For 2 MHz Clock: 354 μs 26

27 Solution 1 code (1435 μs) is faster than Solution 2 (354 μs), but Solution 1 won t fit in the given memory chip Sometimes speed is critical and the system will have enough space for a program that is otherwise size inefficient Solution 1 does not work because it does not fit in the available memory chip Solution 2 does fit in the memory chip Conclusion: Whenever working with a list of data (arrays), use a loop 27

It translates (converts) assembly language to machine code.

Assemblers 1 It translates (converts) assembly language to machine code. Example: LDAA $0180 Uses an instruction set manual: Tests/Final Exam. B6 01 80 Use software: Like the IDE in the Lab. 2 Assembler: