1. Instruction Set Architecture
CHAPTER 6
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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. Where do you go from here (CSE241): 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. Mother Board
Contains packages of integrated circuit chips (IC chips) including a processor, cache (several), memory (DRAM), connections for IO devices (networks, disks)
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5. 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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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.
Lets look at MIPS: Microprocessor without Interlocked Pipelined Stages
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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 (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. Logical operations Shift left Shift right Bit-wise AND Bit-wise OR
Example: sll $t2,$so,4
Reg t2 = $so << 4
Shift right
Example: srl $t2,$so,4
Reg t2 = $so >> 4
Bit-wise AND
Example: and $t0,$t1, $t2
Reg t0 = reg t1 & reg t2
Bit-wise OR
Example: or $t0,$t1,$t2
Reg $t0 = $t1 | $t2

16. Instructions for Selection (if..else)
If (i == j) then f = g + h; else f = g – h;
bne $s3,$s4, else
add $s0,$s1,$s2
j done
else: sub $s0,$s1,$s2
done:

17. Instructions for Iteration (while)
while (save[i] == k)
i = i + 1;
Let i be in reg $s3
Let k be in reg $s5
Let $t1 have the address of Save array element
Loop: sll $t1,$s3,2
add $t1,$t1$s6
lw $t0,0($t1)
bne $t0,$s5,Exit
addi $s3,$s3,1
j Loop

18. Compiling C procedures
int leaf_example (int g, int h, int i, int j) { int f; f = (g + h) – (i + j); return f; } How do you pass the parameters? How does compiler transport the parameters?

19. Passing Parameters/arguments
Special registers for arguments: a0, a1, a2, a3 Save temp register on the stack Perform operations And return value Restore values stored on the stack Jump back to return address

20. Summary MIPS operands MIPS memory MIPS Assembly language
MIPS instructions type and formats
And of course, the four design principles.
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