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Programming for the 68K

Programming for the 68K. Instructor: Rob Nash Readings: Chap 7-9 Midterm next Monday! Review this Wednesday You will see code again…. Hello World. Lets declare a message to display to the world org $600 *start of data HMSG DC.B 'Hello!' EOM DC.B 0 * eom end $400.

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Programming for the 68K

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  1. Programming for the 68K • Instructor: Rob Nash • Readings: Chap 7-9 • Midterm next Monday! • Review this Wednesday • You will see code again…

  2. Hello World • Lets declare a message to display to the world org $600 *start of data HMSG DC.B 'Hello!' EOM DC.B 0 *eom end $400

  3. Useful Interrupts • A way to get the attention of the CPU in an asynronous manner without polling • Also referred to as Traps, we’ll be interested in the following • Trap #3 *prints out what is pointed to by A3 • Trap #9 *ends the program (convention)

  4. Back to Hello World ORG $400 MOVEA.L #HMSG, A3 TRAP #3 TRAP #9 ORG $600 *start of data HMSG DC.B 'Hello!' DC.B 0 *eom END $400

  5. FTOC Example FTOC SUBI.W #32, D0 MULS #5, D0 DIVS #9, D0 RTS

  6. 68K V.S. 8086 • See p.28, section 2.6

  7. Consider… • If register D3 contains 100030FF and register D4 contains 8E552900, what is the result of • MOVE.W D3, D4? • 8E5530FF • MOVE.L D3, D4? • What if I wanted to move the upper half?

  8. Answer • A word op, so the upper half of D4 remains constant. The lower half will be loaded as follows: 8E5530FF

  9. Query • What will A2 contain after the execution of MOVEA.L A5,A2

  10. Notes • Note that, even though the address registers are 32 bits long, only the lower 24 are used to address memory in the 68K • There are no external address lines for the upper 8 bits! • So, if A0 contains 00007F00, what happens when executing: • MOVE.B (A0), D7 • 0x007F00 holds “0009” • D7 contains 1234FEDC

  11. Answer • D7 contains 1234FE09

  12. Address Register Indirect w PostIncrement • This implements a pop • Use the value, then increment (pop off stack) • The stack grows to lower addresses, shrinks to higher addrs

  13. Predecrement • A5 : 0x00007F00 • D2 : 0x4E4F2000 • 0x007EFF : 3C • 0x007F00 : 09 • 0x007F01 : BA • MOVE.W (A5)+, D2

  14. D2 : 4E4F09BA • Use what A5 is pointing to, then decrement • A5 thus looks up 7F00 • Since a word operation, post decrement by 2 • A5 is now 0x007F02

  15. Address Register Indirect w PreDecrement • This implements a push • Decrement our stack first to make space • Then overwrite this new location with our data • Stacks shrink “downwards”, or to higher addresses

  16. With Displacement • Relocatable code is critically important today, and compilers provide this type of code automatically • Contrast this to code with all absolute memory locations specified – how do we shuffle this around in memory?

  17. Indirect with Predecrement • A2: 0x007F00 • D4: F3052BC9 • //Mem • 0x007EFF : 3C • 0x007F00 : 09 • 0x007F01 : BA • MOVE.B –(A2),D4 • MOVE.B D4, -(A2) *push

  18. D4 : F3052B3C • Answer: • Decrement A2 first, so 0x007EFF • Copy from that location into the lower byte of D4

  19. How the Assembler => Binary • MOVE.B D3, (A5)+ • 00 01 101 011 000 011 => 0x1AC3 • MOVE .B 5 for A5 Mode (An)+ Mode Dn 3 (D3) Dest (ea) Source (ea)

  20. User Input • Trap #3, Address in A3 of string • Trap #15 • D0 holds the “task number” • 2 is for user input, A1 points to this for you • 0 is for terminal output • A1 holds the memory addr of the start of string • D1 holds the length of the str to print

  21. Query • What does this code look like in ASM? • adder( int x, int y) { • int z; • z = x + y; • }

  22. Answer • *assuming D0 holds x, D1 holds y, D2 is z • adder AND.W #0, D2 ADD.W D0, D2 ADD.W D1, D2 RTS

  23. Sample Problem #1: What If? • We wanted to compare two lists of bytes? • A1 points to the first list • A2 points to the second list • Each list is 5 elements long • D0 holds 0 if different, 1 if identical. • For 2 reapers!!

  24. Query • What does the asm look like for the following? • branch( inta, intb) { • if( a > b ) { • b = a • } else { • a = b • } • }

  25. Sample Problem #2 • Write an ASM module that defines two (3x3) matrices and sums up the first row • Data section:

  26. Query • What does the following code look like in ASM? • for(int a = 0; a < 10; a++ ) { • nop • }

  27. Sample Problem #3 • Implement the overriding feature of inheritance at the ASM level • Data Section holds a VTable • A table of function pointers • To override a function, we’ll need to update its table entry • I’ll give you the table in A0 • The offset of the function to override in D0 (in longs) • The address of the new function in A1

  28. HW problem • What is a good HW problem for us to consider?

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