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ECE 3430 – Introduction to Microcomputer Systems University of Colorado at Colorado Springs

ECE 3430 – Introduction to Microcomputer Systems University of Colorado at Colorado Springs. Lecture #7 Agenda Today Decoding Machine Code Logical Instructions (AND, OR, exclusive OR, …) Arithmetic Instructions (addition, subtraction, multiplication, division)

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ECE 3430 – Introduction to Microcomputer Systems University of Colorado at Colorado Springs

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  1. ECE 3430 – Introduction to Microcomputer SystemsUniversity of Colorado at Colorado Springs Lecture #7 Agenda Today • Decoding Machine Code • Logical Instructions (AND, OR, exclusive OR, …) • Arithmetic Instructions (addition, subtraction, multiplication, division) • Performing integer multiplication and division • More Instruction Execution and Timing ECE 3430 – Intro to Microcomputer Systems Fall 2014

  2. Decoding Machine Code - The process of creating the assembly language (including mnemonics) for the instruction codes and operands scattered throughout program memory (the machine code). - If we know the contents of the program memory, we can reverse-engineer the code to assembly level. - There is a single mapping from machine code to assembly code and vice-versa. - FYI: It is not always possible to reverse-engineer machine code to a source code form more abstract than assembly (i.e. to the C or C++ source code level). Information is combined and dismissed during the “higher level language” to assembly translation. There can be an infinite number of ways to write higher level code that ultimately produces the same machine code. Debuggers for these higher level languages rely on other constructs (such as symbol files) to map a particular machine instruction back to a single line of source code. ECE 3430 – Intro to Microcomputer Systems Fall 2014

  3. Decoding Machine Code *.s43 file Program Memory Program Execution: • The instruction code (IC) contains all the information about size and addressing mode of the instruction. • The IC inherently tells you where the next IC is stored in memory—and so on. • Just as assemblers can assemble your assembly code into machine code, inverse-assemblers can take machine code and produce the original assembly code. • Why would you want to do this? 1) To verify that the cross-assembler did what you expected. 2) If the address & data busses are “probe-able”, you can use a logic analyzer to track the program and find our where the programs are and what they are doing (nicer logic analyzers have “built-in” inverse assemblers for many common CPUs). • . • The MSP430 Instruction Set Encodings link tells you the IC for a given mnemonic and addressing mode. You can use this to look up the opposite. cross-assembler mov.b &2,R6 0x4256 0x0002 *.s43 file mov.b &2,R6 inverse-assembler (machine code decoding) ECE 3430 – Intro to Microcomputer Systems Fall 2014

  4. Decoding Machine Code Decode the following: 0x4075 0x00331) First we look up 0x4075 in the MSP430 Instruction Set Encodings. The first nybble “4” indicates a mov operation. 2) According to the datasheet format, the next 4 bits represents the source register (0 in this case). 3) The next 4 bits indicate a combination of source and destination addressing mode and the size of the operation: Ad(0):B/W(0):As(1:0) 7 = 0111b Ad = 0  register direct B/W = 1  8-bit (byte) As = 11  register indirect autoincrement 4) The last 4 bits indicate the destination register (5 in this case). At this point we know the source line was: mov.b @R0+,R5 R0 = PC so  mov.b @PC+,R5  mov.b #0033h,R5 ECE 3430 – Intro to Microcomputer Systems Fall 2014

  5. Decoding Machine Code Decode the following: 0x1A 0x42 0x7A 0x001) First we look up 0x421A in the MSP430 Instruction Set Encodings. The first nybble “4” indicates a mov operation. 2) According to the datasheet format, the next 4 bits represents the source register (2 in this case). 3) The next 4 bits indicate a combination of source and destination addressing mode and the size of the operation: Ad(0):B/W(0):As(1:0) 1 = 0001b Ad = 0  register direct B/W = 0  16-bit (word) As = 01  register indexed 4) The last 4 bits indicate the destination register (10 in this case). At this point we know the source line was: mov.w ?(R2),R10 R2 = SR = CG1 (zero generator) so offset ‘?’ is absolute. Operand provides ‘?’ as 16-bit.  mov.w 007Ah(R2),R10  mov.b &007Ah,R10 ECE 3430 – Intro to Microcomputer Systems Fall 2014

  6. Decoding Machine Code Decode the following: 0x56 0x40 0x1B 0x1E1) First we look up 0x4056 in the MSP430 Instruction Set Encodings. The first nybble “4” indicates a mov operation. 2) According to the datasheet format, the next 4 bits represents the source register (0 in this case). 3) The next 4 bits indicate a combination of source and destination addressing mode and the size of the operation: Ad(0):B/W(0):As(1:0) 5 = 0101b Ad = 0  register direct B/W = 1  8-bit (byte) As = 01  register indexed 4) The last 4 bits indicate the destination register (6 in this case). At this point we know the source line was: mov.b ?(R0),R6 R0 = PC. Operand provides ‘?’ as 16-bit.  mov.b 1E1Bh(PC),R6  mov.b <label at PC+1E1Bh>,R6 ECE 3430 – Intro to Microcomputer Systems Fall 2014

  7. Bit-Wise Manipulation of Memory The MSP430 is a byte-addressable architecture. This means that each address targets a byte stored in memory. So what if you want to change one bit in memory (without effecting the other bits in that byte)? Apply boolean logic: AND: Force a bit to zero or no change. OR: Force a bit to one or no change. Exclusive OR (XOR): Toggle a bit to opposite state or no change. ECE 3430 – Intro to Microcomputer Systems Fall 2014

  8. Bit-Wise Manipulation of Memory There is no explicit OR instruction in the MSP430 instruction set. Note there is an AND and XOR instruction. To perform an OR operation, use the BIS instruction. ECE 3430 – Intro to Microcomputer Systems Fall 2014

  9. Bit-Wise Manipulation of Memory The AND, BIS, and XOR instructions conceptually form a logic gate between each corresponding bit of the operand and the value already stored in the register. The source and destination obey by this logic: destination = source <op> destination Like the mov operation, the source can use any of the 4 fundamental addressing modes—while the destination can only use register direct and register indexed. ECE 3430 – Intro to Microcomputer Systems Fall 2014

  10. Bit-Wise Manipulation of Memory The AND, BIS, and XOR instructions conceptually form a logic gate between each corresponding bit of the operand and the value already stored in the register. These logical operations can be performed directly on memory if desired. and <src>,<dest>  bitwise AND 7 6 5 4 3 2 1 0 7 6 5 4 3 2 1 0 x x x x x x x x bis <src>,<dest>  bitwise OR xor <src>,<dest> bitwise exclusive OR Logic Gate 8 or 16 of these (depending on size specified) ECE 3430 – Intro to Microcomputer Systems Fall 2014

  11. Bit-Wise Manipulation of Memory These logical instructions are useful for setting, clearing, or toggling bits within a byte:OR w/ 1 to set mov.b &P1OUT,R4 bis.b #00010000b,R4 mov.b R4,&P1OUT  will force P1.4 to ‘1’ AND w/ 0 to clear and.b #11101111b,&P1OUT  will force P1.4 to ‘0’ XOR w/ 1 to toggle xor.b #00010000b,&P1OUT  will toggle the state of P1.4 ECE 3430 – Intro to Microcomputer Systems Fall 2014

  12. Bit-Wise Manipulation of Memory AND operation also useful for bit masking. For example, compare the state of only the least-significant 4 bits of a byte  mask off the upper 4 bits. XOR is useful for a quick, efficient comparison for equality. A single XOR gate is a 1-bit digital comparator.  xor.w R4,R5 (R5 will be zero if the contents of R4 and R5 are the same) ECE 3430 – Intro to Microcomputer Systems Fall 2014

  13. Carry Flag and the MSP430 The carry flag (C-flag) in the SR always represents a carry condition in this architecture. During addition, it represents whether or not an unsigned overflow condition has occurred. During subtraction, it represents whether or not a larger value has been subtracted from a smaller value when the two values (of equal precision) as both considered unsigned. A “carry” and “borrow” condition are always inverses. When the carry flag is 0 following a subtraction, a “borrow” has occurred (smaller – larger). When the carry flag is 1 following a subtraction, a “borrow” has not occurred (larger - smaller). ECE 3430 – Intro to Microcomputer Systems Fall 2014

  14. Arithmetic Instructions: The ADD Instruction ADD – binary add instruction. ADDC – binary add instruction with carry. ADC – add carry to destination. Can natively do 8 and 16-bit binary addition. Greater than 16-bit requires multi-precision arithmetic. Example add instructions: add.w R5,R6 ; R5 + R6  R6 add.w 4(R4),R7 ; M(R4 + 4) + R7  R7 addc.w @R7,R8 ; M(R7) + R8 + C-flag  R8 adc.b R7 ; R7 + C-flag  R7 ECE 3430 – Intro to Microcomputer Systems Fall 2014

  15. Arithmetic Instructions: The ADD Instruction ADD Instruction Execution –Ex) add.w 4(R2),R7 (2 words, 3 clock cycles)Step 1: PC is put on the address bus (E000h) and a READ is initiated. The data (5417h) is put on the data bus by the memory and grabbed by the CPU. The control unit decodes the instruction code and recognizes the instruction, src and dest addressing modes, and size of operation. The PC is incremented by 2 (PC = PC + 2 = E002h).Step 2: PC is put on the address bus (E002h) and a READ is initiated. The data (0004h) is put on the data bus by memory and grabbed by the CPU. The PC is incremented (PC = PC + 2 = E004h).Step 3: The address (R2+0004h) is put on the address bus (by uC) and the data at that address is put on the data bus (by memory). This value is added to R7 and the result latched into R7. Memory R2+0004h data - - E000h 17h E001h 54h E002h 04h E003h 00h ECE 3430 – Intro to Microcomputer Systems Fall 2014

  16. Arithmetic Instructions: The ADD Instruction Write 8-bit code to add M(0010h) and M(0011h) and store in M(0012h) –mov.b 0010h,R10 ; upper 8 bits of R10 are cleared add.b 0011h,R10 ; lower 8 bits used in add, upper 8 stay 0 mov.b R10,0012h ; writes only the lower 8 bits out to memory The above would add the contents of 0010h and 0011h together. The sum resides in R10 until we store the result out to memory. In this case the result is stored in memory location 0012h. Even though the data bus is 16-bits wide, only 8 bits are accepted by the memory in this case. In other words, memory location 0013h is unaffected. Could one replace the .b with .w to turn everything into 16-bit addition? Any problems doing that? ECE 3430 – Intro to Microcomputer Systems Fall 2014

  17. Arithmetic Instructions: The SUB Instruction SUB –binary subtraction. SUBC – binary subtraction with borrow. SBC – subtract borrow from destination. Example subtract instructions: sub.w R5,R6 ; R6 - R5  R6 sub.w 4(R4),R7 ; R7 - M(R4 + 4)  R7 subc.w @R7,R8 ; R8 - M(R7) - ~C-flag  R8 sbc.b R7 ; R7 - ~C-flag  R7 See the MSP430 Instruction Set Summary link to see how the MSP430 actually performs these operations. They re-word these operations in terms of addition—but conceptually accomplishes the same thing. ECE 3430 – Intro to Microcomputer Systems Fall 2014

  18. Arithmetic Instructions: The SUB Instruction What happens if the result is negative?The “N” bit in the SR is set. What happens if two’s compliment overflow occurs?The “V” bit in the SR is set. What happens if the result is zero?  The “Z” bit in the SR is set. What happens if a larger unsigned value is subtracted from a smaller unsigned value?  The “C” bit is cleared. It is set on larger – smaller. ECE 3430 – Intro to Microcomputer Systems Fall 2014

  19. Arithmetic Instructions: Multiplication Some CPU architectures have support for an integer multiply instruction. Some have unsigned integer only and some have signed and unsigned integer multiply instructions. Some have floating point and some don’t. Most uC CPUs do not have floating point support. Some have 8-by-8, some 16-by-16, some 16-by-8, et cetera. None of the MSP430 CPUs have native support for multiplication. Some versions of the MSP430 have a hardware multiplier (dedicated hardware outside of the CPU) to pull off multiplication. Our MSP430 in the lab does not have a hardware multiplier—so multiplication must be done in software. ECE 3430 – Intro to Microcomputer Systems Fall 2014

  20. Arithmetic Instructions: Multiplication ECE 3430 – Intro to Microcomputer Systems Fall 2014

  21. Arithmetic Instructions: Multiplication ECE 3430 – Intro to Microcomputer Systems Fall 2014

  22. Arithmetic Instructions: Multiplication ECE 3430 – Intro to Microcomputer Systems Fall 2014

  23. Example of a Hardware Multiplier ECE 3430 – Intro to Microcomputer Systems Fall 2014

  24. Multiplication Shortcuts ECE 3430 – Intro to Microcomputer Systems Fall 2014

  25. Arithmetic Instructions: Division Similarly, the MSP430 does not have native support for integer division. Division algorithms for all MSP430 varieties require a software algorithm. ECE 3430 – Intro to Microcomputer Systems Fall 2014

  26. Arithmetic Instructions: Division ECE 3430 – Intro to Microcomputer Systems Fall 2014

  27. Arithmetic Instructions: Division ECE 3430 – Intro to Microcomputer Systems Fall 2014

  28. Arithmetic Instructions: Multiplication and Division Much better multiplication and division algorithms have been designed for computers. Horner’s Method:Provides ways to perform calculations using fewer operations (polynomial math). Booth’s Algorithm:Provides algorithms to work directly with two’s complement integers without conversion to unsigned first. ECE 3430 – Intro to Microcomputer Systems Fall 2014

  29. Arithmetic Instructions: Other Considerations When does the carry bit matter?If the native precision of the operation is not big enough, the carry bit is used to perform multi-precision arithmetic. Discussed later. Is this addition/subtraction operation signed or unsigned?Doesn’t matter. Addition and subtraction works for both. How the rest of the program uses the results of the add/subtract operation differs depending on whether you are dealing with signed or unsigned values (it is all in how you interpret the binary). Multiplication and division algorithms, on the other hand, do need to account for whether the values are signed or unsigned. Architectures with integer multiplication and division instructions are designed to work with either signed or unsigned numbers (sometimes an instruction for each). ECE 3430 – Intro to Microcomputer Systems Fall 2014

  30. Instruction Execution/Timing Timing- An electronic computer is a synchronous system.- This means that all circuitry is synchronous to a clock (memory, CPU, I/O).- The master clock is derived either from an external crystal oscillator on an on-chip oscillator. - The on-chip oscillator can be either the VLO (very low-frequency oscillator) or DCO (digitally-controlled oscillator). - External crystal oscillators provide accuracy when needed.- The master clock (MCLK) can optionally be divided by 2, 4, or 8. - The cycle time (time for each CPU cycle to execute) is 1/(MCLK freq).- Crystal frequencies vary from in the kHz to many MHz and beyond. Ex) How long does the instruction “mov.w #34,R6” take to execute? - First we calculate the number of read and write cycles. - The sum of the read and write cycles is the total number of cycles. - We multiply the total number of cycles by the MCLK period. Execution Time = (# of cycles)(execution time per clock) ECE 3430 – Intro to Microcomputer Systems Fall 2014

  31. Instruction Execution/Timing Instruction Execution: mov.w R4, &1234h (Cycles: 4 cycles; Size: 2 words) (Machine Code = 4482h 1234h)Step 1 - READ IC from PC location. - CPU drives PC value on the address bus. - 4482h is output from memory onto the data bus, CPU receives 4482h. - PC is incremented twice (PC = PC + 2).Step 2 - READ operand from PC location. - CPU drives PC value on the address bus. - 1234h is output from memory onto the data bus, CPU receives 1234h - PC is incremented twice (PC = PC + 2).Step 3 - READ M(1234h) even though not needed for move. - 1234h is driven onto the address bus by CPU - Memory responds with data at memory location 1234h. Step 4 - WRITE contents of R4 to memory location 1234h. - 1234h is driven onto the address bus by CPU. - CPU drives the data bus with contents of R4. ECE 3430 – Intro to Microcomputer Systems Fall 2014

  32. Code Execution - Example Add the contents of M(0010h) and M(0011h) and store result in M(0012h):mov.b 0010h, 0012h ; M(0012h)  M(0010h) add.b 0011h, 0012h ; M(0012h)  M(0011h) + M(0012h) 1) How much memory does this code consume?mov.b 0010h, 0012h ; 3 words add.b 0011h, 0012h ; 3 words6 words = 12 bytes 2) How long does this code take to execute?mov.b 0010h, 0012h ; 6 cycles add.b 0011h, 0012h ; 6 cycles 12 cycles ECE 3430 – Intro to Microcomputer Systems Fall 2014

  33. Code Execution – Alternate (Better) Add the contents of M(0010h) and M(0011h) and store result in M(0012h):mov.b 0010h, R10 ; upper 8 bits of R10 are cleared add.b 0011h, R10 ; lower 8 bits used in add, upper 8 stay 0 mov.b R10, 0012h ; writes only the lower 8 bits out to memory 1) How much memory does this code consume?mov.b 0010h, R10 ; 2 words add.b 0011h, R10 ; 2 words mov.b R10, 0012h ; 2 words6 words = 12 bytes 2) How long does this code take to execute?mov.b 0010h, R10 ; 3 cycles add.b 0011h, R10 ; 3 cycles mov.b R10, 0012h ; 4 cycles 10 cycles ECE 3430 – Intro to Microcomputer Systems Fall 2014

  34. Code Execution – A Little Can Equal A Lot Eliminating even a single cycle from within certain areas of code could make a big difference in performance and, hence, power consumption! What is the improvement (from previous slides)?Size: (12-10) x 100 ~= 17%12 If it executes only once (or very infrequently), 17% is not noticeable. If, on the other hand, the code were executed thousands of times, the speed improvement would become very apparent (if executed in a loop for example)! ECE 3430 – Intro to Microcomputer Systems Fall 2014

  35. End of Exam 1 Material This is the end of the exam 1 material. Prepare questions for next lecture. ECE 3430 – Intro to Microcomputer Systems Fall 2014

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