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CDA4150 Lecture 4

CDA4150 Lecture 4. Topics Covered. Addressing Modes Multiplication of binary integers. Addressing Modes . Direct Immediate Indirect Register Direct Register Indirect Register Indirect plus Offset. Direct Addressing . Used for manipulating an absolute address.

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CDA4150 Lecture 4

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  1. CDA4150 Lecture 4

  2. Topics Covered • Addressing Modes • Multiplication of binary integers

  3. Addressing Modes • Direct • Immediate • Indirect • Register Direct • Register Indirect • Register Indirect plus Offset

  4. Direct Addressing Used for manipulating an absolute address. Example: LOAD R1, <ADDR> will load the contents of <ADDR> into R1 Memory … LOAD 3,100 5 Registers 100 0 … 1 2 200 3 5 CPU …

  5. Immediate Addressing Used when dealing with constants. Example: LOAD R1, #value will load value into R1. LOAD 3,#25 Registers 0 1 2 3 25 CPU

  6. MAR Indirect Addressing Typically used for accessing a value through a pointer. Example: LOAD R1,I,<ADDR> Contents of <ADDR> stores another address in which the contents will be loaded into R1. Memory … LOAD 1,I,500 800 800 Registers 500 0 … 1 18 2 800 18 3 CPU …

  7. Register Direct Addressing Used to copy the contents of one register into another. Example: COPY R2,R1 or MOVE R2,R1 will copy contents of R1 into R2. COPY 3,1 Registers 0 1 25 2 3 25 CPU

  8. MAR Register Indirect Addressing Typically used for accessing a list of consecutive memory locations. Example: LOAD R2,<R1> will load the contents of address stored in R1 into R2 Memory … LOAD 2,1 200 Registers 100 0 … 1 200 2 18 200 18 3 CPU …

  9. Registers 0 + 1 2 3 Register Indirect Addressing plus Offset Typically used when accessing array and structures. Example: LOAD R2,R1,offset will load the contents of address stored in R1+offset into R2 LOAD 2,1,500 Memory … 100 100 18 … CPU 18 600 …

  10. Multiplication of two binary integers • Consider multiplying two 4 bit integers in a 4 bit machine 1011 multiplicand (MDR) X 1101 multiplier (Q) 1000 1111 result (R) Result is 8 bits resulting in overflow condition!

  11. Design Representation R (result) C A Q (multiplier) MDR (multiplicand) ALU Carry over

  12. Multiplication of two binary integers R C Q3 Q2 Q1 Q0 MDR (multiplicand) ALU Carry over

  13. Initial Conditions 0 0 0 0 0 1 1 0 1 1 0 1 1 ALU Carry over

  14. Steps to Follow Initial Step: A = to MDR • Test Q0 for 1 or 0 • a. If Q0 = 1; C, A=A+MDR Move bits to right.. (shift_right C-A-Q) b. If Q0 = 0; (shift_right C-A-Q) • Do step 1 until all bits in Q are tested

  15. Step-by-Step • Fill initial values 0 0 0 0 0 1 1 0 1 1 0 1 1 A C Q0 = 1 MDR

  16. Step-by-Step • Set A = MDR 0 1 0 1 1 1 1 0 1 1 0 1 1 A C Q0 = 1 MDR

  17. Q0 = 1 • C = A+MDR 0 1 0 1 1 1 1 0 1 1 0 1 1 MDR A Q0 = 1 0 0 1 0 1 1 1 1 0 1 0 1 1 MDR A Q0 = 1 • Shift bits to the right Step-by-Step

  18. Q0 = 0 0 0 1 0 1 1 1 1 0 1 0 1 1 MDR A Q0 = 1 0 0 0 1 0 1 1 1 1 1 0 1 1 MDR A Q0 = 1 • Shift bits to the right Step-by-Step

  19. Q1 = 1 • C = A + MDR 0 1 1 0 1 1 1 1 1 1 0 1 1 MDR A Q0 = 1 0 0 1 1 0 1 1 1 1 1 0 1 1 MDR A Q0 = 1 • Shift bits to the right Step-by-Step

  20. Q1 = 1 • C = A + MDR 1 0 0 0 1 1 1 1 1 1 0 1 1 MDR A Q0 = 1 0 1 0 0 0 1 1 1 1 1 0 1 1 MDR A Q0 = 1 • Shift bits to the right Step-by-Step

  21. 0 1 0 0 0 1 1 1 1 1 0 1 1 MDR A Q0 = 1 R Step-by-Step • All Bits in Q tested • R = result 0 1 0 0 1 1 1 1 1 1 0 1 1 MDR A Q0 = 1

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