CS 241 – Week 2 Tutorial Solutions - Student.cs.uwaterloo.ca

CS 241 – Week 2 Tutorial Solutions
Assembly Language Programming
Spring 2013
1 Problem 0 - Convert MIPS to ML
1. add $3, $3, $0
The format for an add instruction is:
0000 00ss ssst tttt dddd d000 0010 0000
in this case, d = 3(00011), s = 3(00011), t = 0(00000). So the final 32-bit binary is:
0000 00
}
00
{{
011
}
0
}
0000
{{ }
0001
} {{
1
}
000 0010 0000
$s $t $d
The Hexadecimal representation is: 0x00601820.
2. slt $3, $20, $30
The format for a slt instruction is:
0000 00ss ssst tttt dddd d000 0010 1010
in this case, d = 3(00011), s = 20(10100), t = 30(11110). So the final 32-bit binary is:
0000 00
}
10
{{
100
}
1
}
1110
{{ }
0001
} {{
1
}
000 0010 1010
$s $t $d
The Hexadecimal representation is: 0x029E182A.
3. beq $0, $0, -4
The format for an beq instruction is:
0001 00ss ssst tttt iiii iiii iiii iiii
in this case, s = 0(00000), t = 0(00000), i = -4(1111 1111 1111 1100). So the final 32-bit binary is:
0001 00
}
00
{{
000
}
0
}
0000
{{ }
1111
}
1111
{{
1111 1100
}
$s $t
i
The Hexadecimal representation is: 0x1000FFFC.
2 Problem 1 - How to write a MIPS loop
Sol(a) ;set up constants
lis $5
.word 1
add $3, $0, $0
slt $2, $0, $1
beq $2, $0, 3
;$2 = 0 => 0 !< n
add $3, $1, $3
;result += n
sub $1, $1, $5 ;n -= 1
1

eq $0, $0, -5
jr $31
Sol(b) ;$2 = i
;$1 = n
;$3 = result
;$5 = 1
;set up constants
lis $5
.word 1
;set $3 to 0, $2 to 1
add $3, $0, $0
add $2, $5, $0
slt $6, $1, $2 ;i n)
beq $6, $5, 3 ;$6 = 1 => (n < i)
add $3, $3, $2
add $2, $2, $5
beq $0, $0, -5
;result += i
;i++
jr $31
Sol(c) ;set up constants
lis $5
.word 1
add $3, $0, $0
add $3, $1, $3
;result += n
sub $1, $1, $5 ;n -= 1
bne $1, $0, -3
;$1 != 0, then loop
jr $31
3 Problem 2 - More looping
We want to compute the nth fibonacci number. Assume the value of n is in $1 and that the nth fibonacci
number fits in a 32-bit integer. Compute the nth fibonacci number and store it in $3, then return.
;find the nth fib number
;$1 = n
;$2 = F_n-1
;$3 = F_n
;$5 = 1
;high level inspiration
;int fib(int n) {
; int a = 1;
2

; int b = 0;
; while (n != 0) {
; int temp = a + b;
; a = b;
; b = temp;
; n = n -1
; }
; return b;
;}
lis $2
.word 1 ; a = 1
add $5, $2, $0 ; $5 = 1
add $3, $0, $0 ; b = 0
beq $1, $0, 5
; n != 0, goes into loop
add $4, $2, $3
; temp = a + b
add $2, $3, $0
; a = b
add $3, $4, $0
; b = temp
sub $1, $1, $5 ; n = n - 1
beq $0, $0, -6
jr $31
; return to top of loop
; return b
3