Classifying Instruction Set Architectures Stack Register- Register Register – Memory Accumulator Memory – Memory

Stack The code C=A+B Push A Push B Add Pop C

Accumulator The Code C=A+B Load A Add B Store C

Register-Memory The Code C=A+B Load R1,A Add R3,R1,B Store R3,C

Load-Store/ Register-register Code C=A+B Load R1,A Load R2,B Add R3,R1,R2 Store R3,C Virtually every new architecture design after 1980 uses load-store register architecture!

ANOTHER EXAMPLE A = B + C B = A + C D = A – B Assumptions: Op Code is 8 bit represented by O, Address is 16-bit represented by A. All registers and data is 32-bit.

EXAMPLE: Accumulator Instruc Comments Size of Operand Code MemoryUsage Bytes MemoryUsage Load B accumulator  B O+A 3 4 Add C accumulator  B + C Store A store B + C in [A] Add C accumulator  A + C Store B store A + C in B Negate negate accumulator O 1 Add A accumulator  B + A Store D store A – B in D Total = 22 28

Example: Mem to Mem Instruction Comments Size of Operand Code Bytes Memory Usage add A, B, C ; MEM[A] = MEM[B] + MEM[C] O+A+A+A 7 12 add B, A, C ; MEM[B] = MEM[A] + MEM[C] sub D, A, B ; MEM[D] = MEM[A] – MEM[B] Total = 21 36

Load-Store Instruc Comments Size of Operand Code Memory Usage 4 3 R= Reg Field (4-bit) Code ≈ Bytes Memory Usage Bytes LW R1, B R1  MEM[B] O+A+R = 28bits 4 LW R2, C R2  MEM[C] ADD R3, R1, R2 R3  B + C O+R+R+R = 20bits 3 SW A, R3 MEM[A] = B + C ADD R1, R3, R2 R1  A + C SW B, R1 MEM[B] = A + C SUB R4, R3, R1 R4  A – B SW D, R4 MEM[D] = A – B Total = 29 20

STACK Instruc Comments Size of Operand Code Memory Usage Push B O+A 3 Bytes Memory Usage Push B ; push B onto stack O+A 3 4 Push C ; push C onto stack Add ; top <- B + C O 1 Pop A ; A = B + C Push A ; push A onto stack ; top <- A + C Pop B ; B = A + C ; push A onto stack ; push B onto stack Sub ; top <- A – B Pop D ; D = A – B Total = 30 36

Example: Reg – Mem Instruction Comments Size of Operand Total = 40 20 R = Reg Spec (1 Byte) Code Bytes Memory Usage Mov ECX, DWORD PTR [C] Mov C to ECX O+A+R 6 4 ADD EBX, DWORD PTR [B] B= B+C Mov EDX, EBX Temp Save O+R 2 Mov EAX,EBX A = B+C Mov DWORD PTR [A], EAX Save A O+A ADD EAX, ECX A+C Mov EBX,EAX B = A+C Mov DWROD PTR [B], EBX Save B SUB EDX, EBX D = A - B Mov DWORD PTR [D], EDX Save D Total = 40 20

Conclusion Architecture Instruction Memory Accesses (in bytes) Data Memory Accesses (In bytes) Total Accumulator 22 28 50 Memory/Memory 21 36 57 Stack 30 66 Load/Store 29 20 49 Reg-Mem(Intel) 40 60

Pros & Cons… Advantages Disadvantages Type Instruction Encoding Code Generation # of Clock Cycles/ Inst. Code Size Register-register Fixed-length Simple Similar Large Register-memory Variable Length Moderate Different Medium Memory-memory Variable-length Complex Large variation Compact Advantages Disadvantages Source: Louisiana state University, ece

Addressing Modes

Addressing Modes

Summery of use of memory addressing mode Displacement 10% 20% 30% 40% 50% Frequency of the addressing mode 0% Tex Spice gcc 1% 6% 16% 24% 3% 11% 17% 43% 39% 32% 55% Memory indirect Scaled Register Indirect Immediate Summary of memory addressing mode

Size of Displacement Percentage of Displacement 45% 40% Percentage of Displacement 35% 30% 25% 20% 15% 10% 5% 0% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Number of Bits needed for Displacement

Use of Immediate Operand

Immediate Addressing Mode- Displacement distribution 45% 40% Percentage of Immediate 35% 30% 25% 20% 15% 10% 5% 0% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Number of Bits needed for Immediate Operand

Instruction Type

Distribution by benchmark size Double Word (64 bits) Word (32 bits) 10% 20% 30% 40% 50% 0% Half word (16bits) Byte (8bits) 16% 6% 29% 26% 5% 1% Distribution of data accesses by size of benchmark 60% 70% 80% Floating point average Integer average

Instructions for Control Flow The Measurements of branch and jump behavior are fairly independent of other measurements and applications. Four types of control flow change: Conditional branches Jumps Procedure calls Procedure returns

Three classes of control flow instructions Conditional Branch 100% Call/return Jump 25% 50% 8% 19% 10% 6% 82% 75% Control Flow instructions into three classes Floating point average Integer average

Addressing modes for flow control instructions Destination address of a control flow must be specified Destination is specified by supplying a displacement that is added to the Program Counter (PC) This type of flow control instructions are called PC-relative. This way code can run independent of where it is loaded; this is called position independence

Addressing modes for flow control instructions Implement returns and indirect jumps when target I not known at compile time. These register indirect jumps are important for other features: Case or switch statements Virtual functions or methods High order functions Dynamically shared libraries

Branch distances in terms of number of instructions

Conditional branch options Conditional Code (CC) register E.g. 80x86,ARM etc. Tests special bit set by ALU operations Advantage Sometimes condition is set free Disadvantage CC is extra state. Condition codes constrain the ordering of instructions since they pass information from one instruction to a branch

Conditional branch options Conditional Register E.g. Alpha, MIPS Tests arbitrary register with the result of a comparison Advantage Simple Disadvantage Uses up register

Conditional branch options Compare and branch E.g. PA-RISC, VAX Compare is part of the branch. Often compare is limited to subset Advantage One instruction rather than two for a branch Disadvantage May be too much work per instruction for pipelined execution

Types of compares in conditional branching Less than Greater than or equal 10% 50% 0% 44% 33% 34% 35% Frequency of comparison types in braches Floating point average Integer average 20% 30% 40% Less than or equal Equal Not Equal 5% 2% 16% 18% 11% Greater than

Encoding an instruction set