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16.317: Microprocessor System Design I

16.317: Microprocessor System Design I. Instructor: Dr. Michael Geiger Spring 2012 Lecture 12: Instruction sequence examples. Lecture outline. Announcements/reminders Graded HW 1 available HW 2 due today Lab 1 due 2/29 Amanda’s OH: M/W 2:30-5, Tuesday 5-7 in Ball 407 Exam 1: Friday, 2/24

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16.317: Microprocessor System Design I

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  1. 16.317: Microprocessor System Design I Instructor: Dr. Michael Geiger Spring 2012 Lecture 12: Instruction sequence examples

  2. Lecture outline • Announcements/reminders • Graded HW 1 available • HW 2 due today • Lab 1 due 2/29 • Amanda’s OH: M/W 2:30-5, Tuesday 5-7 in Ball 407 • Exam 1: Friday, 2/24 • Review lecture Wednesday, 2/22 • Will be allowed calculator, one 8.5”x11” sheet of notes • No class Monday, 2/20 • Lecture outline • Review: shift/rotate instructions • Today: more practice with sequences of instructions Microprocessors I: Lecture 12

  3. Review: shift instructions • Basic shift instructions • Move value by <amt> bits; add 0s to left or right • CF = last bit shifted out • SHL <src>, <amt>: Move <src> to left • SAL exactly the same • SHR <src>, <amt>: Move <src> to right • Arithmetic right shift • Move value right by <amt> bits • Copy sign bit to fill remaining bits • CF = last bit shifted out • SAR <src>, <amt> Microprocessors I: Lecture 12

  4. Review: rotate instructions • Rotate instructions: bits that are shifted out one side are shifted back in other side • ROL <src>, <amt> or ROR <src>, <amt> • CF = last bit rotated • Rotate through carry instructions • CF acts as “extra” bit that is part of value being rotated • RCL <src>, <amt> Microprocessors I: Lecture 12

  5. Rotate example • Given AL = 43H, CL = 04H, and CF = 0, show the state of AL after each instruction in the sequence below: • ROR AL, 2 • ROL AL, CL • RCR AL, 3 • RCL AL, 4 Microprocessors I: Lecture 12

  6. Solution • Initially, AL = 43H = 010000112 • ROR AL, 2 • AL = 01000011 rotated right by 2 = 11010000 = D0H • CF = last bit rotated in = 1 • ROL AL, CL • AL = 11010000 rotated left by 4 = 00001101 = 0DH • CF = last bit rotated in = 1 Microprocessors I: Lecture 12

  7. Solution (cont.) • RCR AL, 3 • (AL,CF) = 00001101 1 rotated right by 3 = 01100001 1 • CF = 1, AL = 011000012 = 61H • RCL AL, 4 • (CF,AL) = 1 01100001 rotated left by 4 = 0 00011011 • CF = 0, AL = 000110112 = 1BH Microprocessors I: Lecture 12

  8. Instruction sequence examples: initial state • EAX: 00000000H • EBX: 00000000H • ECX: 00000000H • EDX: 00000000H • ESI: 00000000H • EDI: 00000000H • EBP: 00000000H • ESP: 00000000H • DS: 1000H • SS: 0000H Microprocessors I: Lecture 12

  9. Sequence Example 1 • Show the results of the following sequence • LDS SI, [0100H] • LSS DI, [SI] • MOV SP, FFFEH • MOV BP, SP Microprocessors I: Lecture 12

  10. Sequence Example 1: solution • LDS SI, [0100H] • SI = word from DS:0100H = word from10100H = 0004H • DS = word from DS:0102H = word from 10102H = 1010H • LSS DI, [SI] • DI = word from DS:SI = word from 10104H = 0012H • SS = word from DS:SI+2 = word from 10106H = 4020H • MOV SP, FFFEH • MOV BP, SP • BP = SP = FFFEH Microprocessors I: Lecture 12

  11. Instruction sequence examples: state after sequence 1 • EAX: 00000000H • EBX: 00000000H • ECX: 00000000H • EDX: 00000000H • ESI: 00000004H • EDI: 00000012H • EBP: 0000FFFEH • ESP: 0000FFFEH • DS: 1010H • SS: 4020H Microprocessors I: Lecture 12

  12. Sequence Example 2 • Show the results of the following sequence • MOV CX, [SI+0004H] • MOV AX, [SI+000CH] • SHL AX, CL • ADD AX, [SI+000CH] • ADD AX, [SI+000CH] • MOV [DI], AX Microprocessors I: Lecture 12

  13. Sequence Example 2: solution • MOV CX, [SI+0004H] • CX = word at DS:SI+0004H = word at 10108H = 0002H • MOV AX, [SI+000CH] • AX = word at DS:SI+000CH = word at 10110H = 001EH • SHL AX, CL • AX = AX << (CL) = 001EH << 2 = 0078H Microprocessors I: Lecture 12

  14. Sequence Example 2: solution • ADD AX, [SI+000CH] • AX = AX + word at DS:SI+000CH = 0078H + 001EH = 0096H • ADD AX, [SI+000CH] • AX = AX + word at DS:SI+000CH = 0096H + 001EH = 00B4H • MOV [DI], AX • Word at DS:DI = AX • DS:DI = 10100H + 0012H = 10112H • Word at 10112H = 00B4H Microprocessors I: Lecture 12

  15. Instruction sequence examples: state after sequence 2 • EAX: 000000B4H • EBX: 00000000H • ECX: 00000002H • EDX: 00000000H • ESI: 00000004H • EDI: 00000012H • EBP: 0000FFFEH • ESP: 0000FFFEH • DS: 1010H • SS: 4020H Microprocessors I: Lecture 12

  16. Instruction sequence examples: state before sequence 3 • Some time later, assume: • EAX: 000000B4H • EBX: 00000000H • ECX: 00000002H • EDX: 00000000H • ESI: 00000004H • EDI: 00000012H • EBP: 0000FFF8H • ESP: 0000FFF4H • DS: 1010H • SS: 4020H • Memory from 10100-10116 unchanged Microprocessors I: Lecture 12

  17. Sequence Example 3 • Show the results of the following sequence • MOV SI,SS:[BP+0004H] • MOV DI,SS:[BP+0006H] • MOV AX,[SI] • XCHG [DI],AX • XCHG DI,SI • MOV [DI],AX Microprocessors I: Lecture 12

  18. Sequence Example 3: solution • MOV SI,SS:[BP+0004H] • Note: if BP is base register, SS is implied • SI = word at SS:BP+0004H = word at 501FCH = 000AH • MOV DI,SS:[BP+0006H] • DI = word at SS:BP+0006H = word at 501FEH = 000CH • MOV AX,[SI] • AX = word at DS:SI = word at 1010A = 1600H • XCHG [DI],AX • Swap contents of AX with word at DS:DI (address 1010C) • AX = 0317H, word at 1010C = 1600H • XCHG DI,SI • Swap DI and SI  DI = 000AH, SI = 000CH • MOV [DI],AX • Word at DS:DI = contents of AX • Word at 1010AH = 0317H Microprocessors I: Lecture 12

  19. Instruction sequence examples: state after sequence 3 • EAX: 00000317H • EBX: 00000000H • ECX: 00000002H • EDX: 00000000H • ESI: 0000000CH • EDI: 0000000AH • EBP: 0000FFF8H • ESP: 0000FFF4H • DS: 1010H • SS: 4020H Microprocessors I: Lecture 12

  20. Next time • Exam 1 Preview (on Wednesday) • Remember, no class Monday Microprocessors I: Lecture 12

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