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9/29: Lecture Topics

• Conditional branch instructions• Unconditional jump instructions• Hexadecimal/Binary/Decimal• Instruction encoding

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Mailing List and Computer Labs

• CSE410 mailing list– Subscribe if you weren’t automatically

or you use a different account• Send email to

majordomo@cs.washington.edu with ‘subscribe cse410’ in the body

• Computer accounts– MSCC Lab in basement of

Communications building– Username = passwd =

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Conditional Branch

yes

no

?

...

...

...

...

A change of the flow of control

of the program that depends on

a condition

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Control flow in C

• Conditional branch constructs:– while, do while loops– for loops

• Unconditional jump constructs:– goto– switch statements

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Branching Instructions

• beq• bne• other real instructions: b, bgez, bgtz, more...

• other pseudoinstructions: beqz, bge, bgt, ble, blt, more...

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Pseudoinstructions

• One-to-one mapping between assembly and machine language not strictly true

• Assembler can do some of the work for you

• Example: slt is a real instruction; blt is a pseudoinstruction

• move is a pseudoinstruction

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Labels in MIPS

• MIPS allows text tags– a name for a place to branch to

• Under the covers, the assembler converts the text tag into an address

loop: add $t0, $t0, $t0 #

bne $t0, $t1, loop #

sw $t2, 0($t3) #

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Building a while loop in MIPS

i=0;while(i<100) { i++;}

Idea: translate

into assembly.

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How about a for loop?

• One answer: translate from for to while, then use while loop conventions

• This is typically how compilers handle loops

for(i=0; i<N; i++) { do_stuff(i);}

i=0;while(i<N) { do_stuff(i); i++;}

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Example C Program

int array[100];

void main() {

int i;

for(i=0; i<100; i++)

array[i] = i;

}

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An Assembly Version

main: move $t0, $0 #

la $t1, array #

start: bge $t0, 100, exit #

sw $t0, 0($t1) #

addi $t0, $t0, 1 #

addi $t1, $t1, 4 #

j start #

exit: j $ra #

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Unconditional Jump

• (Mostly) the same as branch• No choice about it; just go to the

label

• How are branch and jump different?– we’ll see when we get to instruction

encoding

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Binary Numbers

• Instead of 0-9 we can only use 0 and 1

• Also known as base-2

• Counting to 10ten in binary 0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010

• Binary arithmetic 1 1011+101010101

11001001- 1010110

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Binary -> Decimal

• Converting from Base-2 to Base-1012 = 20 = 110

102 = 21 = 210

1002 = 22 = 410

10002 = 23 = 810

11102 = 1*23 + 1*22 + 1*21 + 0*20

= 8+4+2+0 = 1410

1010112 =

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Decimal -> Binary

• Repeatedly divide by 2 until you reach 0

• Reverse the order of the remainders5010

50/2 = 25 remainder 025/2 = 12 remainder 112/2 = 6 remainder 0 6/2 = 3 remainder 0 3/2 = 1 remainder 1 1/2 = 0 remainder 1

1100102

7510

75/2 = remainder /2 = remainder /2 = remainder /2 = remainder /2 = remainder /2 = remainder /2 = remainder

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Hexadecimal Numbers

• Instead of 0-9 we use 0-15 • 0-15 is confusing so we use 0-9, A-

F• Also known as base-16

• Counting to 18ten in hex 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, 10, 11, 12

• Hex arithmetic

1 1 2E97+A2B3 D14A

A3AD38B993CA93C930B-A3AD38B993CA93C930A

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Hexadecimal <-> Binary

• Each four binary digits represents one hexadecimal digit

• Can quickly convert either way– 0x2A07 = 0010 1010 0000 0111– 1111 0011 0101 1011 =

00002 = 016

00012 = 116

00102 = 216

00112 = 316

01002 = 416

01012 = 516

01102 = 616

01112 = 716

10002 = 816

10012 = 916

10102 = A16

10112 = B16

11002 = C16

11012 = D16

11102 = E16

11112 = F16

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Hexadecimal -> Decimal

• Converting from Base-16 to Base-10116 = 160 = 110

1016 = 161 = 1610

10016 = 162 = 25610

100016 = 163 = 153610

A32 = A*161 + 3*160

= 10*161 + 3*160 = 160 + 3 = 163

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Decimal -> Hexadecimal

• Repeatedly divide by 16 until you reach 0

• Reverse the order of the remainders50010

500/16 = 31 remainder 4 31/16 = 1 remainder 15 (F) 1/16 = 0 remainder 1

1F416

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What you need to know

• Be able to convert to/from small decimal numbers from/to binary or hex

• Be able to/from convert any binary number from/to a hex number

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Stored Program

• So far, we’ve seen that data comes from memory

• It turns out that the program comes from memory also

• Each instruction in the program has an address

• Von Neumann computer (p. 33)

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Program Layout (p. 160)

Address0

0x00400000

0x10000000

0x10008000

0x7fffffff

Reserved

Text

Static data

Dynamic data and stack

Program instructions

Global variables

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The Text Segment

• Let’s zoom in on the text segment:

0x00400000

0x10000000

Text

0x004000000x004000040x00400008

add $t0, $t1, $t2

sub $t0, $t0, $t3

lw $s1, 4($t0)

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Jump Register

• Now we’re ready for the jr instruction

• Syntax:

• Effect: jump to the instruction at the address in $t0

jr $t0

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Another Look at Labels

• Labels are text tags

• The assembler translates them into addresses and does search-and-replace

loop: add $t0, $t0, $t0

bne $t0, $t1, loop

sw $t2, 0($t3)

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Instruction Encoding

• How does the assembler translate each instruction into bits?

add $t0, $s1, $s2

00000010001100100100000000100000

assembler

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Fields

• Split the 32-bit instruction into fields:

• op+funct: Which instruction?• rs, rt: Register source operands• rd: Register destination operand

32 bits

op rs rt rd shamt funct

6 bits 5 bits 5 bits 5 bits 5 bits 6 bits

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• Let’s translate:

• Opcode for add is 000000 (p. A-55)• The funct for add is 100000• The code for $t0 is 01000 (p. A-23)• The code for $s1 is 10001• The code for $s2 is 10010

Encoding an add instruction

add $t0, $s1, $s2

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Instruction Formats

• The 32 bits can be split up more than one way

• For add, it was 6-5-5-5-5-6– This is called R-format

• For lw, we use 6-5-5-16 instead– This is called I-format

32 bits

op rs rt address

6 bits 5 bits 5 bits 16 bits

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Translating into I-format

• The opcode for lw is 100011• The code for $t0 is 01000• The code for $t1 is 01001• Binary for 100 is

lw $t0, 100($t1)

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Branch vs. Jump

• Branch instructions use the I-format– The destination address is 16 bits– You can only branch ±32KB away

• Jump instructions use another format called the J-format– The destination address is 26 bits– You can jump up to ±32MB away

• What’s the tradeoff?

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What to Take Away

• I don’t care if you:– Know how many bits are in each field– Know what the opcodes are– Know what the codes for the registers

are

• I do care that you:– Know what fields are– Know why we need more than one

instruction format