Machine and Assembly Language Byte Addressable Memory

(In class notes)

L2

Programs are data. Programs are formed by instructions.

Registers: very fast memory. RAM: we use ram to refer to memory.

L3

slt $d,$s, $t-SetLessThan.ifs<tsetsthevalueofregister$d to be 1, other wise sets is it to be 0. - Notes: needs to know if it’s signed or unsigned. Only works if both are the same.

EX: Negates the value in register $1provideditispositiveslt$2, $1,$0: if 1<0, sets register 2 to 1. $2isgoingtobebne$2, $0,1:skipifitisnegative.?sub$1, $0,$1: substract dollar 1 from 0, that takes the complement? jr $31 Idea: Register 2 is 0 if register 1 is non-negative. Branch if register 2 is not zero. Otherwise negate register 1.

EX: Put absolute value of register $1intoregister$2 name file .asm vim abs.asm add $2,$1, $0:movedollar1intodollar2slt$8, $2,$0: $8issetsto1if$0 < $8;$8 = $2<0:iflessthan0thanitsnegativesoyoudo{n}^{\prime }tskipthenextline.beq$8, $0:if$8 is 0, then you’ve done previous, skip 1 instruction sub $2,$0, $2:negatebysubstractingby0jr$31: number is positive FIX beq $8,$0, ifend sub $2$0, $2ifend:jr$32

LOOPING Adss together all even numbers from 1 to 20 incluseve and store answer in register $3lis$2 .word 20 lis $1.word2add$3, $0,$0: sets the result to 0 becuase we want to 0, start the loop at 20 and go down. (less code) add $3,$3, $2;line-3sub$2, $2,$2; line -2 bne $2,$0, -3; line -1 from here, if doesn’t reach 0 we go back to line -3, and come down again jr $31

Use label for us to add new instructions

EX: sample has the address 0x4, which is the location of add $1,$0, $0

lis $2
.word 20
lis $1
.word 2
add $3 $0, $0
top:
	add $3 $3 $2
	sub $2 $2, $1
	bne $2 $0, top
jr $31

label below condition vs above is a loop careful with where you check and put the label…