1. Chapter 11 Instruction Sets: Addressing Modes and Formats
Ren-Jie Wang 王仁傑, Ph.D.
資訊館7F辦公室
上課時間: 星期三 13:25—16:10

2. Addressing Modes 定址模式 Immediate Addressing 立即定址 Direct Addressing 直接定址
Indirect Addressing 間接定址
Register Addressing 暫存器定址
Register Indirect Addressing 暫存間接器定址
Displacement Addressing 位移定址
Relative Addressing 相對定址
Base-Register Addressing 基底暫存器定址
Indexed Addressing 索引定址
Stack Addressing 堆疊定址 2

3. Operand is part of instruction Operand = address field e.g. ADD 5
Immediate Addressing
Operand is part of instruction
Operand = address field
e.g. ADD 5
Add 5 to contents of accumulator ( AC←AC＋5 )
5 is operand
No memory reference to fetch data
Fast
Limited range （why?） 3

4. Immediate Addressing Diagram
Instruction
Opcode
Operand
運算碼
(指令)
運算元
(資料) 4

5. Direct Addressing Address field contains address of operand
Effective address （有效位址）EA = address field A
e.g. ADD A
Add contents of cell A to accumulator ( AC←AC＋A )
Look in memory at address A for operand
Single memory reference to access data
No additional calculations to work out effective address
Limited address space （why?） 5

6. Direct Addressing Diagram
Instruction
Opcode
Address A
Memory
ADD A
Operand 6

7. Indirect Addressing (1)
Memory cell pointed to by address field contains the address of (pointer to) the operand
EA = (A)
Look in A, find address (A) and look there for operand
e.g. ADD (A)
Add contents of cell pointed to by contents of A to accumulator
(A) : Content of location A or register A 7

8. Indirect Addressing (2)
Large address space
2N where N = word length
May be nested, multilevel, cascaded
e.g. EA = (((A)))
Draw the diagram yourself
Multiple memory accesses to find operand
Hence slower 8

9. Indirect Addressing Diagram
Instruction
Opcode
Address A
Memory
ADD
(A)
Pointer to operand
Operand 9

10. Indirect Addressing Diagram
Instruction
Opcode
Address A
Memory
ADD
((A))
Pointer to operand
Pointer to operand
Operand 9

11. Register Addressing (1)
Operand is held in register named in address filed
EA = R
Limited number of registers
Very small address field needed
Shorter instructions
Faster instruction fetch 10

12. Register Addressing (2)
No memory access
Very fast execution
Very limited address space
Multiple registers helps performance
Requires good assembly programming or compiler writing
C programming
register int a;
c.f. Direct addressing 11

13. Register Addressing Diagram
Instruction
Opcode
Register Address R
Registers
Operand 12

14. Register Indirect Addressing
C.f. indirect addressing
EA = (R)
Operand is in memory cell pointed to by contents of register R
Large address space (2N)
One fewer memory access than indirect addressing 13

15. Register Indirect Addressing Diagram
Instruction
Opcode
Register Address R
Memory
ADD
(R)
Registers
Pointer to Operand
Operand 14

16. Displacement Addressing
EA = A + (R)
Address field hold two values
A = base value 基底
R = register that holds displacement 位移
or vice versa 反之亦然 15

17. Displacement Addressing Diagram
Instruction
Opcode
Register R
Address A
Memory
Registers
Pointer to Operand +
Operand 16

18. A version of displacement addressing R = Program counter, PC
Relative Addressing
A version of displacement addressing
R = Program counter, PC
EA = A + (PC) PC-relative
i.e. get operand from A cells from current location pointed to by PC
c.f locality of reference & cache usage 17

19. + Relative Addressing Instruction Opcode Address A Memory PC Operand
Pointer to Operand
Operand

20. Base-Register Addressing
A holds displacement
R holds pointer to base address
R may be explicit or implicit 外顯或隱含
e.g. segment registers in 80x86 18

21. Good for accessing arrays
Indexed Addressing
A = base
R = displacement
EA = A + (R)
Good for accessing arrays
(R) = (R) +1 19

22. Combinations Pre-index 前索引定址模式 (先更新索引,再存取資料)
然後基底暫存器的值再更新成新的位址
EA = (A+(R))
ARM指令: LDR R0, [R1, #4]! ;R0:=[R1+4], R1:=R1+4
Post-index 後索引定址模式 (先存取資料,後更新索引)
基底暫存器的值會先用來當所要存取的記憶體位址
存取資料後，基底暫存器的值再更新成新的位址
EA = (A) + (R)
ARM指令: LDR R0, [R1], #4 ;R0:=[R1], R1:=R1+4 20

23. Pre-index: EA = (A+(R))
Instruction
Opcode
Register R
Address A
Memory
Base
Registers
Displacement +
Pointer to Operand
Operand

24. Post-index: EA = (A) + (R)
Instruction
Address A
Opcode
Register R
Memory
Registers
Base
Displacement +
Operand

25. Operand is (implicitly) on top of stack e.g.
Stack Addressing
Operand is (implicitly) on top of stack
e.g.
ADD Pop top two items from stack and add
Textbook p. 398, problems #11.13 21

27. IA-32 (Pentium 4) Registers
Streaming SIMD Extensions
MMX: SIMD instruction set

28. purposes in different

30. Pentium Addressing Modes
Virtual or effective address is offset into segment
Starting address plus offset gives linear address
This goes through page translation if paging enabled
12 addressing modes available
Immediate
Register operand
Displacement
Base
Base with displacement
Scaled index with displacement
Base with index and displacement
Base scaled index with displacement
Relative

34. Pentium Addressing Mode Calculation

36. Instruction Formats
11.3
Layout of bits in an instruction
Includes opcode
Includes (implicit or explicit) operand(s)
Usually more than one instruction format in an instruction set

37. Affected by and affects:
Instruction Length
Affected by and affects:
Memory size
Memory organization
Bus structure
CPU complexity
CPU speed
Trade off between powerful instruction repertoire and saving space

38. Allocation of Bits
Number of addressing modes
Number of operands
Register versus memory
Number of register sets
Address range
Address granularity

39. Pentium Instruction Format

40. In-class exercise
Given the following memory values and a one-address machine with an accumulator, what values do the following instructions load into the accumulator? （Textbook p. 397, problems #11.1）
Word 20 contains 40
Word 30 contains 50
Word 40 contains 60
Word 50 contains 70
LOAD IMMEDIATE 20
LOAD DIRECT 20
LOAD INDIRECT 20
LOAD IMMEDIATE 30
LOAD DIRECT 30
LOAD INDIRECT 30 20 40 30 50 60 70

41. In-class exercise
Assume a stack-oriented processor that includes the stack operations PUSH and POP. Arithmetic operations automatically involve the top one or two stack elements. Begin with an empty stack. What stack elements remain after the following instructions are executed? （Textbook p. 398, problems #11.13）
PUSH 4
PUSH 7
PUSH 8
ADD
PUSH 10
SUB
MUL