1. Instruction Set Architecture
Chapter 2
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2. Topics for Discussion Instruction set architecture
Stored program computer
Interface between software and hardware
Instructions
Design principles
Instructions, registers, and memory
Simple instructions types and formats
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3. Computer Organization
Is the study of major components of a modern digital computer, their organization and assembly, and the architecture and inner workings of these components.
It also deals with design principles for a good performance.
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4. Central Processing Unit (CPU)
Example: Intel 80386 80486Pentium
Main components of a CPU are datapath and control unit
Datapath is the component of the processor that performs (arithmetic) operations
Control is the component of the processor that commands the datapath, memory , IO device according to instruction of the program
Cache provides but fast memory that acts as a buffer for slower /larger memory outside the chip.
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5. Importance of ISA
How are high level language (HLL) translated to the language of the hardware.
How does hardware execute instructions?
What is the interface between hardware and software?
How does the software instruct (or control) the hardware to perform the needed function?
What determines the performance of a program?
How can a program improve performance?
Original program, its translation, effectiveness of the hardware, other special techniques such as pipelining
What techniques can by hardware designers to improve performance?
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6. Instruction set Architecture
An important abstraction between hardware and software.
Lets discuss this concept.
Computer operation is historically called an instruction.
Instructions stored similar to data in a memory give rise to an important foundational concept called the stored program computer.
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7. Consider a C language statement:
C to MIPS instruction
Consider a C language statement:
f = (g + h) – ( i + j)
Compile
add t0, g, h
add t1,i, j
sub f,t0,t1
Design principle 1: simplicity favors regularity
In the above example: all instructions have 3 operands
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8. Register set Where do the data get stored in the CPU?
Named locations called registers? How many? Typical small compared to memory sizes.
Registers: MIPS-32 has 32 register
Denoted by s0, s1, etc. $s0, $s5
Temporary registers are denoted by $t0, $t1
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9. C to MIPS instruction (Take 2 with registers)
Consider a C language statement:
f = (g + h) – ( i + j)
Compile
add $t0, $s1, $s2
add $t1,$s3,$s4
sub $s0,$t0,$t1
Design principle 2: Smaller is faster
Memory available as registers is 32 in number
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10. Memory Operations
Data and instructions are stored in memory outside the CPU.
Data is loaded from memory and stored in memory.
Load word (lw)
Store word (sw)
32 resgiters
230 words or 232 addressable locations or bytes
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11. C language to Memory instructions
g = h + A[8]
Compile
lw $t0, 32($s3)
add $s1, $s2, $t0
sw $t0,48($s3)
Base register concept: base register is $s3 and
Offset of 32 for 8 words and offset of 48 for 12 words
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12. Instruction Types add and sub lw and sw
Now lets see how we can deal with a constant value data.
Consider C language statement: x = x +4
Too complex:
lw $t0, AddrConst($s1)
add $s3,$s3,$t0
Instead how about:
addi $s3,$s3,4
Design principle 3: Make the common case fast. Example “addi” instead of add an constant from memory.
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13. Instruction format –R typeOp
6 bits RS
5 bits Rt Rd
Shamt
funct
Can we use the same format for addi and add? Then we will
Have only 11 bit constant
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14. Instruction format – I typeOp
6 bits Rs
5 bits Rt
Constant or address
16 bits
Design principle 4: good design demands good compromise;
Keep instruction length same needing different formats ; I and R type are
examples
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15. Summary Figure 2.1 very nicely summarizes all we learned so far.
MIPS operands
MIPS memory
MIPS Assembly language
MIPS instructions type and formats
And of course, the four design principles.
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