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ECE 265 – Lecture 6

ECE 265 – Lecture 6. The M68HC11 Basic Instruction Set Basic Arithmetic Instructions. Lecture Overview. The M68HC11 Basic Instruction Set The basic mathematical instructions REF: Chapter 3 and the appendix that details the instructions. Arithmetic Instructions.

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ECE 265 – Lecture 6

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  1. ECE265 ECE 265 – Lecture 6 The M68HC11 Basic Instruction Set Basic Arithmetic Instructions

  2. Lecture Overview • The M68HC11 Basic Instruction Set • The basic mathematical instructions • REF: Chapter 3 and the appendix that details the instructions. ECE265

  3. Arithmetic Instructions • These instructions are used to add, subtract, compare, increment, decrement, 2’s complement, test, and decimal adjust data. • Both 8-bit and 16-bit operations are possible. • It is possible to write code that allows data of higher precision to be operated upon. This extended precision requires the user to write code to support the higher precision. • Architecture also supports Binary Coded Decimal (BCD) operations. ECE265

  4. Add instructions • Add instructions – note the addressing mode ECE265

  5. Add operations • Operation: • ABA – add the A and B accumulators – Result in A • ABX, ABY – add B to the specified index register • ADCA,ADCB – add with carry to A/B • ADDA,ADDB – add the contents of memory to A/B • ADC and these set the H CC bit to accommodate BCD arithmetic • ADDD – a 16-bit addition using memory • Description: 2’s complement binary addition • CC effects: N X V C and H as noted ECE265

  6. The overflow flag • What exactly is overflow? Does that mean you exceeded the binary range represent-able? ECE265

  7. More arithemetic instructions • Decimal adjust, increment and decrement, and • Two’s complement. ECE265

  8. Decimal Adjust DAA • Description: Adjusts the result in A to a valid BCD number. • Consider adding the BCD for $99 + $22 in Accum A • Both $99 and $22 are valid BCD numbers. • Adding these in binary gives a result of $BB with no half carry or carry. Also, $B is not a valid BCD digit. • Executing DAA will have the following effect. • For the lsd: BCD of $9 + $2= $B will give a $1 and a half carry. • For the most significant digit you have $9 + $2=$B adjusted $B is $1 + hc = $2 and a final carry out. • So here, accumulator A is adjusted to $21 and both C & H are set • CC effects: N Z and C H are corrected • Forms: DAA and only inherent mode ECE265

  9. Increment and Decrement • Description: Instructions that increment or decrement a memory location or either the A or B accumulator (Inherent mode). Note that there is no instruction to increment or decrement the D accumulator. (and C is not affected) • CC effects: N Z V • Forms: INCA INCB INC (opr) • DECA DECB DEC(opt) • Just add 1 or subtract 1 (2’s complement) ECE265

  10. Two complement operations • Description: Replace the contents of the location with its two’s complement value • CC effects: N Z V and C • The C bit will be set in all cases except when the contents are $00 • Forms: NEGA NEGB NEG (opr) ECE265

  11. Two’s Complement • How do you get the two’s complement? • Two simple algorithms – consider 00001001(dec 9) ECE265

  12. Two’s Complement • How do you get the two’s complement? • Two simple algorithms – consider 00001001(dec 9) • 1. Take the 1’s complement and add 1 • Ones complement is 11110110 + 00000001 = 11110111 • Which represents -9 ECE265

  13. Two’s Complement • How do you get the two’s complement? • Two simple algorithms – consider 00001001(dec 9) • 1. Take the 1’s complement and add 1 • Ones complement is 11110110 + 00000001 = 11110111 • Which represents -9 • 2. Starting with the lsb keep all binary digits until you come to the 1st 1. Keep that 1. Invert all the more significant binary digits. • Easy to see on the above example. • Now consider 0000 1110  11110010 (14 and -14) ECE265

  14. Two’s complement numbers • 0111 7 • 0110 6 • 0101 5 • 0100 4 • 0011 3 • 0010 2 • 0001 1 • 0000 0 • 1111 -1 • 1110 -2 • 1101 -3 • 1100 -4 • 1011 -5 • 1010 -6 • 1001 -7 • 1000 -8 For 4-bits ECE265

  15. Additional Arithmetic Instr. • A few more instructions, this time for binary subtraction ECE265

  16. Binary subtraction • Consider the following example of 7 – 5 = 2 • 0111 • -0101 • 0010 Direct and very easy to see • Now throw in a borrow - 6 – 5 = 1 • 0110 • -0101 but right off see that we need to subtract 1 from 0 • so borrow from the next binary position and get • 0102 (and yes 2 is not a binary digit – but this is for illustration) • -0101 (and it is the weight when a digit is moved right) • 0001 ECE265

  17. More binary subtraction • Consider 12 – 3 = 9 • 1100 • -0011 and we again have borrows ECE265

  18. More binary subtraction • Consider 12 – 3 = 9 • 1100 • -0011 and we again have borrows • 1020 after 1st step of borrow • -0011 but we still need a borrow ECE265

  19. More binary subtraction • Consider 12 – 3 = 9 • 1100 • -0011 and we again have borrows • 1020 after 1st step of borrow • -0011 but we still need a borrow • 1012 after 2nd step of borrow • -0011 • 1001 which is the binary for 9 ECE265

  20. How about a 2’s complement result • An example to get a 2’s complement result, i.e., a negative number result. • Consider 3 – 5 • 0011 • - 0101 • Borrow in =1 2011  1211 • - 0101 • giving 1110 which is the 2’s • complement for -2 • And here would also have a CC carry bit of 1 to indicate a borrow. ECE265

  21. The subtract instruction • Description: Subtract the contents of two locations • CC effects: N Z V and C • C is set if contents of 2nd operand is larger • Forms: SBA – subtract accumulator B from A  A • SUBA (opr) – subtract memory location • SUBB (opr) contents from accumulator • SUBD (opr) - 16 bit operation • SBCA (opr), SBCB (opr) – subtract with • carry ECE265

  22. Compare Instructions • Instruction to compare two values and set condition codes. ECE265

  23. Compare Instructions • Description: compare the data at two locations and set the CC bits. The data is not altered. (Compare instructions perform a subtraction to update the bits of the CC register.) • CC effects: N V C Z • Forms: CBA • CPD (opr) CMPA (opr) and CMPB (opr) • Addressing modes: Note the difference in the mnemonic for register compare of D, A, or B to memory. ECE265

  24. Compare instruction example • Comparison of a subtract versus a compare. A subtract instruction does alter one of the operands which is the destination. • Note that only the affected • bit of the CCR is shown. ECE265

  25. Compare example problem • PROBLEM: What programming steps are needed to compare the data at address $1031 with a set point value of $50 using accumulator A. Note that this address is one of the A/D result registers, i.e., where the result of an A/D conversion is stored. ECE265

  26. Compare example problem • PROBLEM: What programming steps are needed to compare the data at address $1031 with a set point value of $50 using accumulator A. Note that this address is one of the A/D result registers, i.e., where the result of an A/D conversion is stored. • LDDA #$50 Load the set point into Reg A • CMPA $1031 Compare A to memory (A-M) • Results will indicate if A=M : Z=1 • A>M : N=0 Z=0 • A<M : N=1 Z=0 ECE265

  27. Another example • If the data at address $1031 in the previous example was $45 which bits in the CC register are set or cleared? ECE265

  28. Test Instruction • The test instructions • Allows testing of values for positive, negative, or zero values. • The instruction subtracts $00 from the location, setting the CC register bits. The contents of the location are not modified. ECE265

  29. Multiply and Divide instructions • The processor can perform and 8-bit by 8-bit multiply and two forms of divide. The forms for divide are integer and fractional. ECE265

  30. The multiply • Example 3.12 from text: What programming steps are needed to multiply the data at location $D500 with the data at location $D510. The result is to be stored at $D520 and $D521. • LDAA $D500 Load value #1 into A accum • LDAB $D510 Load value #2 into B accum • MUL A x B -> D • STD $D520 Store result (16 bits) ECE265

  31. Another multiply example • Figure 3.5 ECE265

  32. Binary Multiplication • Multiply in the previous example • $FF = 1111 1111 255 • $14 = 0001 0100 20 • 11 1111 1100 5100 • 1111 1111 0000 • 1 0011 1110 1100 ($13EC) • Is this 5100? • 4096 +512+256 +128+64+32 +8+4 • 4096 +768 +128+96 +12 • 4864 +140 +96 = 4864 + 236 = 5100  ECE265

  33. The Divide instruction • 2 divide instructions • IDIV Binary integer division • Used when D is larger than X • Divide D/X -> X with the remainder going into D • FDIV Binary fractional division • Used when X is larger than D • Divide D/X -> X with the remainder (or continuation of the fractional part going into D ECE265

  34. Division Examples • Integer and Fractional examples ECE265

  35. Lecture summary • Have covered • Cover data transfer instructions in Lecture 5 • In this lecture went over the basic arithmetic instructions • Add • Subtract • Increment and Decrement • Testing data • Multiply and Divide ECE265

  36. Assignment • Problems Chapter 3 page 87 • Problem 13 • Problem 14 • Problem 17 ECE265

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