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Physics 413 Chapter 2 : HCS 12 Assembly Programming

Physics 413 Chapter 2 : HCS 12 Assembly Programming. My First Assembly Program. Program adds two 16-bit numbers stored at $1000 - $1001 and $1002 - $1003 and stores the sum at $1010 - $1011 org $2000 ldd $1000 addd $1002 std $1010 end. Assembler Directives and more …. org end equ db

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Physics 413 Chapter 2 : HCS 12 Assembly Programming

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  1. Physics 413Chapter 2: HCS 12 Assembly Programming

  2. My First Assembly Program Program adds two 16-bit numbers stored at $1000 - $1001 and $1002 - $1003 and stores the sum at $1010 - $1011 org $2000 ldd $1000 addd $1002 std $1010 end

  3. Assembler Directives and more … org end equ db fcc Examples alpha equ 1.7 string fcc “hello!” array db $3, $8, $23, $11, $57, $12, $47

  4. DAA DAA Decimal Adjust Accumulator Converts binary addition of BCD numbers into BCD format What will the contents of A be with & without DAA ? LDAA # $ 2386 23 ADDA # $ 488B 48 DAA 19 SWI3F

  5. DAA LDAA # $ 23 ADDA # $ 48 DAA SWI Without DAA the contents of A = 6B With DAA the contents of A = 71

  6. Multiplication and Division mulA * B stored in A:B emul D * Y stored in Y:D ediv 32-bit number Y:D divided by 16-bit X and quotient stored in Y and remainder in D

  7. emacs Multiply and Accumulate Multiplies two 16-bit numbers and adds a 32-bit number. 16-bit numbers are pointed to by index registers X and Y. Result is stored where the 32-bit constant was stored! 16-bit number takes up two memory locations 32-bit number takes up 4 memory locations. Useful in signal processing and other computational problems Rewrite ax2 +bx +c as x(ax+b) +c for emacs format (Mx . Mx+1 )*(My . My+1 ) + (M . M +3)  (M . M +3)

  8. A Fork in the Road LDAB # $ 13 here : ADDA # $ 24 DEC B BNE here SWI BNEis the new instruction. Branch if not equal to zero.

  9. Branches Galore MnemonicDescription of Criteria BCC Branch if Carry Clear BCSBranch if Carry Set BEQ Branch if Equal to Zero BNE Branch if Not Equal to Zero BRA Branch Always Note:Familiarize yourself with other related instructions like DBNE,JMP,BSR, and JSR

  10. Dare to Compare ! here: LDAA # $ E3 CMPA $ 50 BNE here SWI Temperature sensor reading is stored at memory location 0050 and compared with a danger limit of E3 .

  11. Bit Condition Branch Instruction again brclr $1000,$84,again ldd $70 The mask is $84. The bits to be checked are bit 2 and bit 7. The contents of memory location $1000 will be checked and when bits 2 and 7 are both 0 the instruction ldd $70 will be executed . If both bits 2 and 7 are not 0 then the program will keep checking by going back to again.

  12. Bit Condition Branch Instruction again brset $1000,$84,again ldd $70 The mask is $84. The bits to be checked are bit 2 and bit 7. The contents of memory location $1000 will be checked and when bits 2 and 7 are both 1 the instruction ldd $70 will be executed . If both bits 2 and 7 are not 1 then the program will keep checking by going back to again.

  13. ASLA ASLA Arithmetic Shift Left Accumulator A. It shifts all the 8 bits of accumulator A one bit to the LEFT. The rightmost bit becomes 0 and the leftmost bit ends up as the carry bit of CCR C 0

  14. Logical AND • What is the outcome of ANDA #$ 4C if accumulator A contains 3 A ?

  15. Logical AND 3A 0011 1010 4C 0100 1100 _________________________ 0000 1000 Hence A will contain 08 after the AND operation

  16. Bit Flipping and Testing • bclr 0,y,$44 clears bits 2 and 6 of the contents of the memory location pointed to by the index register Y. • bset sets the bits designated by the mask. • bita $22 tests bits 1 and 5 of accumulator A and updates the Z and N flags of the CCR but does not change the contents of the accumulator.

  17. Delay Loop DELAY: LDX # $ FFFF AGAIN : DEX BNE AGAIN SWI

  18. Delay Loop Subroutine here:JSRDELAY LDAA # $ E3 CMPA$ 50 BNE here SWI . . . DELAY: LDX # $ FFFF AGAIN : DEX BNE AGAIN RTS

  19. Stack • Stack is the area of RAM pointed to by the 16-bit Stack Pointer (SP) • LDS functions like LDX • LDS #$ 5C42 The number 5C42 is loaded into SP • LDS $ 5C42 Numbers from 5C43 and 5C42 loaded • LDS $ 5C Numbers from 005D and 005C loaded

  20. Push and Pull PSHA A MSP SP - 1 SP PULA SP + 1 SP MSP A

  21. Predict the Outcome! PSHA PSHB PULA PULB

  22. 00D3 00D4 00D5 00D6 Solution Congratulations, if you said the contents of A and B will be swappedand , perhaps more importantly, the value of the stack pointer will be restored to its original value before this program segment was run.

  23. 00D3 00D4 00D5 00D6 Detailed Explanation Suppose that SP was pointing at 00D6 (stack). PUSHA stores A into 00D6. Then PUSHB stores B into 00D5. At this point SP = 00D4. Then PULA pulls 00D5 (which contains B) and stores it into A. Finally, PULB pulls 00D6 (which contains A) and stores it into B. We end up swapping A and B. At this point SP = 00D6, its original value.

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