Computer ArchitectureCPSC 321

E. J. Kim

Overview

• Logical Instructions

• Shifts

Bitwise Operations• Up until now, we’ve done arithmetic (add, sub,addi ),

memory access (lw and sw), and branches and jumps.

• All of these instructions view contents of register as a single quantity (such as a signed or unsigned integer)

° New Perspective: View contents of register as 32 bits rather than as a single 32-bit number

• Since registers are composed of 32 bits, we may want to access individual bits (or groups of bits) rather than the whole.

• Introduce two new classes of instructions:

• Logical Operators

• Shift Instructions

Logical Operators

• Two basic logical operators:• AND: outputs 1 only if both inputs are 1

• OR: outputs 1 if at least one input is 1

Logical Operators

• Truth Table: standard table listing all possible combinations of inputs and resultant output for each

• Truth Table for AND and OR

A B AND OR

0 0

0 1

1 0

1 1

• Two basic logical operators:

– AND: outputs 1 only if both inputs are 1

– OR: outputs 1 if at least one input is 1

Logical Operators

• Instruction Names:

• and, or: Both of these expect the third argument to be a register

• andi, ori: Both of these expect the third argument to be an immediate

• MIPS Logical Operators are all bitwise, meaning that bit 0 of the output is produced by the respective bit 0’s of the inputs, bit 1 by the bit 1’s, etc.

Uses for Logical Operators

• Note that ANDing a bit with 0 produces a 0 at the output while ANDing a bit with 1 produces the original bit.

• This can be used to create a mask.

• Example:

1011 0110 1010 0100 0011 1101 1001 1010

0000 0000 0000 0000 0000 1111 1111 1111

• The result of ANDing these two is:

0000 0000 0000 0000 0000 1101 1001 1010

• The second bit string in the example is called a mask. It is used to isolate the rightmost 12 bits of the first bit string by masking out the rest of the string (e.g. setting it to all 0s).

Uses for Logical Operators

• Thus, the and operator can be used to set certain portions of a bit string to 0s, while leaving the rest alone.

• In particular, if the first bitstring in the above example were in $t0, then the following instruction would mask it:

andi $t0,$t0,0xFFF

• Similarly, note that ORing a bit with 1 produces a 1 at the output while ORing a bit with 0 produces the original bit.

• This can be used to force certain bits of a string to 1s.

• For example, if $t0 contains 0x12345678, then after this instruction:

ori $t0, $t0, 0xFFFF

• $t0 contains 0x1234FFFF (e.g. the high-order 16 bits are untouched, while the low-order 16 bits are forced to 1s).

Shift Instructions (1/3)• Move (shift) all the bits in a word to the left or

right by a number of bits.

• Example: shift right by 8 bits

0001 0010 0011 0100 0101 0110 0111 1000

0000 0000 0001 0010 0011 0100 0101 0110

Example: shift left by 8 bits

0001 0010 0011 0100 0101 0110 0111 1000

0011 0100 0101 0110 0111 1000 0000 0000

Shift Instructions (2/3)• Shift Instruction Syntax:

1 2,3,4

•where

1) operation name

2) register that will receive value

3) first operand (register)

4) shift amount (constant <= 32)

MIPS shift instructions:

1. sll (shift left logical): shifts left and fills emptied bits with 0s

2. srl (shift right logical): shifts right and fills emptied bits with 0s

3. sra (shift right arithmetic): shifts right and fills emptied bits by sign extending

Shift Instructions (3/3)• Example: shift right arith (sra) by 8 bits

0001 0010 0011 0100 0101 0110 0111 1000

0000 0000 0001 0010 0011 0100 0101 0110

Example: shift right arith (sra) by 8 bits

1001 0010 0011 0100 0101 0110 0111 1000

1111 1111 1001 0010 0011 0100 0101 0110

Uses for Shift Instructions (1/4)

• Suppose we want to isolate byte 0 (rightmost 8 bits) of a word in $t0. Simply use:

andi $t0,$t0,0xFF

• Suppose we want to isolate byte 1 (bit 15 to bit 8) of a word in $t0. We can use:

andi $t0,$t0,0xFF00

but then we still need to shift to the right by 8 bits...

Uses for Shift Instructions (2/4)

• Could use instead:

sll $t0,$t0,16

srl $t0,$t0,24

t0

0001 0010 0011 0100 0101 0110 0111 1000

After sll

0101 0110 0111 1000 0000 0000 0000 0000

After srl

0000 0000 0000 0000 0000 0000 0101 0110

Uses for Shift Instructions (3/4)

• In binary:

Multiplying by 2 is same as shifting left by 1:

112 x 102 = 1102

10102 x 102 = 101002

Multiplying by 4 is same as shifting left by 2:

112 x 1002 = 11002

10102 x 1002 = 1010002

Multiplying by 2n is same as shifting left by n

Uses for Shift Instructions (4/4)

• Since shifting maybe faster than multiplication, a good compiler usually notices when C code multiplies by a power of 2 and compiles it to a shift instruction:

a *= 8; (in C)

would compile to:

sll $s0,$s0,3 (in MIPS)

• Likewise, shift right to divide by powers of 2

•remember to use sra

The Story so far…

• We introduced numerous MIPS assembly language instructions.

• We are now familiar with registers and register usage conventions.

• We know how to use system calls, basic I/O• We have learned how the stack works• What is missing?

Practice! Practice! Practice!

What Next?

• We need a more detailed knowledge about the instruction formats to fully appreciate certain restrictions.

• The functional interface is easy to understand, since it is basically familiar procedural programming

• We need to understand how the computer interprets the instruction, so that we can transition to the discussion of the MIPS hardware architecture

Machine Language

• Machine language level programming means that we have to provide the bit encodings for the instructions

• For example, add $t0, $s1, $s2 represents the 32bit string

• 00000010001100100100000000100000• Assembly language mnemonics usually

translate into one instruction• We also have pseudo-instructions that

translate into several instructions

What does that

mean?

Instruction Word Formats

• Register format

• Immediate format

• Jump format

op-code rs rt rd shamt functop-code rs rt rd shamt funct

op-code rs rt immediate valueop-code rs rt immediate value

op-code 26 bit current segment addressop-code 26 bit current segment address

6 5 5 16

6 5 5 5 5 6

6 26

Register Format (R-Format)

• Register format

• op: basic operation of instruction• funct: variant of instruction• rs: first register source operand• rt: second register source operand• rd: register destination operand• shamt: shift amount

op-code rs rt rd shamt functop-code rs rt rd shamt funct

6 5 5 5 5 6

Watson, the case is clear…

• add $t0, $s1, $s2• 00000010001100100100000000100000

• 000000 10001 10010 01000 00000 100000• Operation and function field tell the computer to

perform an addition

• 000000 10001 10010 01000 00000 100000• registers $17, $18 and $8

op-code rs rt rd shamt functop-code rs rt rd shamt funct

6 5 5 5 5 6

0 $zero

1 $at

2 $v0

3 $v1

4 $a0

5 $a1

6 $a2

7 $a3

8 $t0

9 $t1

10 $t2

11 $t3

12 $t4

13 $t5

14 $t6

15 $t7

16 $s0

17 $s1

18 $s2

19 $s3

20 $s4

21 $s5

22 $s6

23 $s7

24 $t8

Number NameValue

Registers pass parameters

to functions

return valuesfrom functions

$s0-$s7 are callee-savedregisters – use theseregisters for valuesthat must be maintainedacross function calls.

$t0-$t7 are caller savedregisters –use these registers in functions

Watson, the case is clear…

• add $t0, $s1, $s2• 00000010001100100100000000100000

• 000000 10001 10010 01000 00000 100000• source registers $s1=$17 and $s2=$18 and

target register $t0=$8

op-code rs rt rd shamt functop-code rs rt rd shamt funct

6 5 5 5 5 6

R-Format Example

• Register format

• (op, funct)=(0,32): add• rs=17: first source operand is $s1• rt=18: second source operand is $s2• Rd=8: register destination is $t0• add $t0, $s1, $s2

0 17 18 8 0 32 0 17 18 8 0 32

6 5 5 5 5 6

Immediate Format (I-Format)

Immediate format

• op determines the instruction (op <> 0) • rs is the source register• rt is the destination register• 16bit immediate value

op rs rt immediate value op rs rt immediate value

6 5 5 16

I-Format Example

Immediate format

• op=8 means addi• rs=29 means source register is $sp• rt=29 means $sp is destination register• immediate value = 4 • addi $sp, $sp, 4

8 29 29 4 8 29 29 4

6 5 5 16

Problem

• The MIPS assembly language has the command andi, an immediate bit-wise and operation

• We can say li $s0, 0xCDEF1234 to load register $s0 with the content 0xCDEF1234

• Why is this strange? • In the immediate format, you can only

load 16 bits, but the constant is 32 bits!

Pseudo-Instructions

• li $s0, 0xCDEF1234 is a pseudo-instruction• It is a convenient shorthand for

lui $at, 0xCDEFori $s0, $at, 0x1234

• The register $at is used here by the assembler; this is the reason why you should not use this register.

Puzzle

• How can we swap the content of two registers, say $s0 and $s1, without accessing other registers or memory?

• Solution: xor $s0, $s0, $s1xor $s1, $s0, $s1xor $s0, $s0, $s1

MIPS Addressing Modes

• Immediate addressing• Register addressing• Base displacement addressing• PC-relative addressing

• address is the sum of the PC and a constant in the instruction

• Pseudo-direct addressing• jump address is 26bits of instruction

concatenated with upper bits of PC

Byte Halfword Word

Registers

Memory

Memory

Word

Memory

Word

Register

Register

1. Immediate addressing

2. Register addressing

3. Base addressing

4. PC-relative addressing

5. Pseudodirect addressing

op rs rt

op rs rt

op rs rt

op

op

rs rt

Address

Address

Address

rd . . . funct

Immediate

PC

PC

+

+

Addressing Modes

• Register Addressing• add $s1, $s2, $s3• $s1 = $s2 + $s3

• Immediate Addressing • addi $s1, $s2, 100• $s1 = $s2 + 100

Addressing Modes

• Base addressing• lw $s1, 100($s2)• $s1 = Memory[$s2+100]

• PC-relative branch• beq $s1, $s2, 25• if ($s1 == $s2) goto PC + 4 + 100

Addressing Modes

• Pseudo-direct addressing• j 1000• goto 1000• concatenate 26bit address with upper

bits of the PC