CHAPTER 6 Instruction Set Architecture 12/7/2015 1

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 12/7/2015 2

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. 12/7/2015 3

Mother Board Contains packages of integrated circuit chips (IC chips) including a processor, cache (several), memory (DRAM), connections for IO devices (networks, disks) 12/7/2015 4

Central Processing Unit (CPU) Example: Intel   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. 12/7/2015 5

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 12/7/2015 6

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 12/7/2015 7

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 12/7/2015 8

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 12/7/2015 9

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 2 30 words or 2 32 addressable locations or bytes 12/7/

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 12/7/

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. 12/7/

Instruction format –R type Op 6 bits RS 5 bits Rt 5 bits Rd 5 bits Shamt 5 bits funct 6 bits Can we use the same format for addi and add? Then we will Have only 11 bit constant 12/7/

Instruction format – I type Op 6 bits Rs 5 bits Rt 5 bits 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 12/7/

Logical operations Shift left  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

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:

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

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?

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

Summary MIPS operands MIPS memory MIPS Assembly language MIPS instructions type and formats And of course, the four design principles. 12/7/