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Machine Code

Machine Code. 1 high level language instruction ≥ 1 machine language instr. 1 assembly language instruction = 1 machine language instr. For 68000, 1 machine language instruction = 1 to 5 words. typically: word 1 operation code, size and type of operands words 2-3

gregfranklin
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Machine Code

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  1. Machine Code 1 high level language instruction ≥ 1 machine language instr. 1 assembly language instruction = 1 machine language instr. For 68000, 1 machine language instruction = 1 to 5 words. typically: • word 1 • operation code, size and type of operands • words 2-3 • immediate operand (1-2 words, if any) • source effective address (1-2 words, if any) • words 4-5 • destination effective address (1-2 words, if any) NOTE: words 2 to 5 are typically not shown in the manual; assumes programmer can figure this out

  2. Machine Code • Immediate data • if byte, low byte of extension word • if word, one extension word • if longword, two extension words:high half : low half • Memory addresses • we always assume xxx.L or longword • Displacement addressing modes • if d(An), d is an extension word (d can be 16 bits) • if d(An,Rm.s), extension word = txxxs000dddddddd (d can only be 8 bits)where t=1(Rm = Ax) or t=0(Rm = Dx) s=1(Rm.L) or s=0(Rm.W) • if An = PC, extension word stores the offset (not responsible for this case)

  3. Machine Code e.g. TRAP #15 e.g. ABCD D0,D1 vs ABCD -(A0),-(A1)

  4. e.g. OR.B D1,D2 vs OR.W #5,D2 vs OR.L NUM,D4

  5. e.g. MOVE.W #$3332,NUM

  6. Instruction Timing Notation: n(r/w) n = number of clock periods for instruction (total) r = number of read cycles for instruction w = number of write cycles for instruction Assumes: t(1 read cycle) = t(1 write cycle) = 4 clock periods n = (r x 4) + (w x 4) + internal processing time e.g. 12(3/0) = 12 clock periods (3 x 4 + 0 x 4 + 0) e.g. 18(3/1) = 18 clock periods (3 x 4 + 1 x 4 + 2)

  7. Instruction Timing e.g. MOVEA.L #R2Z,A1 • look up instruction in table 8-3;where source = #<data> and destination = An • execution time is 12(3/0) • 3 reads from memory • 1st word of machine code • 2 machine code extension words for the immediate data • 0 writes to memory • 0 clock periods for internal processing

  8. Instruction Timing e.g. MOVE.W NUM1,NUM2 • look up instruction in table 8-2; where source = xxx.L and destination = xxx.L(we always assume long for addresses) • execution time is 28(6/1) • 6 reads from memory • 1st word of machine code • 2 extension words for address of NUM1 • 2 extension words for address of NUM2 • fetch of data from NUM1 • 1 write to memory • write data to NUM2 • 0 clock periods for internal processing

  9. Instruction Timing … wait states • if memory is slower than 4 clock periods, longer memory access cycle will insert wait states • must add the wait states to the total instruction time. Assume memory takes an additional 2 wait states (i.e. 2 more clock cycles per memory r/w). if original execution time was 18(3/1) = 18 clock periods (3 x 4 + 1 x 4 + 2) the execution time with the slower memory is ?(3/1) = (3 x (4+2) + 1 x (4+2) + 2) = 26 clock periods

  10. Reading/Expectations Reading: • Section 8: 16-Bit Instruction Execution Times [pages 8-1 to 8-11] in M68000 8-/16-/32-Bit Microprocessors User’s Manual[pdf, 184 p; Motorola] • Add the execution time section to your reference material. Bring it to the exam. • M68000 Machine Code[pdf, 21p; N. Znotinas] • Add the machine code section to your reference material. Bring it to the exam. Expectations: • you are responsible for all material in the reading • you should be able to generate the machine code for any instruction • you should be able to determine the timing for any section of code

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