Final Exam
CMPE 012: Computer Systems and Assembly Language
University of California, Santa Cruz
DO NOT BEGIN UNTIL YOU ARE TOLD TO DO SO.
This exam is closed book and closed notes. Only 4-function calculators are permitted. Answers must be marked on the Scantron form to be graded. All work must be written on the exam.
On the Scantron form, bubble in your name, student ID number, and test form (found in the footer of subsequent pages). In the center of the page write your CruzID, quarter, and exam type. On the back of the page, write the CruzIDs of students sitting to your left and right, and your row and seat number. See below.
On this page, write your last name, first name, CruzID, row and seat numbers, and the CruzIDs of the people to your immediate left and right. Once you are permitted to begin, write your CruzID on all subsequent pages of the exam.
You must sit in your assigned seat. Keep your student or government issued ID on your desk. Brimmed hats must be removed or turned around backwards. Only unmarked water bottles are permitted. Backpacks must be placed at the front of the room or along the walls. Your cell phone must be on a setting where it will not make noise or vibrate.
There are 45 questions on this exam; you only need to answer 42 for full points. The additional three questions (of your choosing) will be counted as extra credit. All questions are multiple choice, and some questions have more than one correct answer. You must mark all correct answers to receive credit for a question. Some true/false questions might list False as answer A and True as answer B. Follow the answers on the exam, NOT the T F notation on the Scantron Form. You will have 120 minutes to complete this exam.
____________________________ ____________________________ ____________________________
Row # Seat #
___________________________________________ Last Name
___________________________________________ CruzID of person to left
CruzID
___________________________________________ First Name
___________________________________________ CruzID of person to right
Bits
1. How many bits are needed to encode one ASCII character? A. 8bits
B. 10bits C. 6bits D. 7bits E. 9bits
2. What is the size of a word in MIPS? Select all that apply. A. 8bytes
B. 32bits C. 8nybbles D. 4bytes E. 32bytes
Binary Arithmetic
3. Perform the following 12-bit two’s complement addition.
0b 0111 1100 0100 + 0b 0000 0001 1100
A. 000011011100 B. 000001010100 C. 011111100001 D. 000010100111 E. 011111100000
4. Which of these 8-bit two’s complement computations has carry out but no overflow? Select all that apply. A. 0x1E + 0x26 = 0x44
B. 0xFA + 0xED = 0xE7
C. 0x0F + 0x85 = 0x94
D. 0x01 + 0x7F = 0x80 E. 0xFF + 0x01 = 0x00
5. A logical right shift and an arithmetic right shift perform the same operation A. True
B. False
CruzID:
@ucsc.edu
CMPE 12 Final Exam – Version A Winter 2019
CMPE 12 Final Exam – Version A Page 1 of 13 Winter 2019
7. What
A. 1235
8. What A.
9. What
is the following 8-bit two’s complement number in signed magnitude form?
B.
C.
D.
E.
-128 to 127 -127 to 128 -128 to 128 -127 to 127
is the following base 9 number in base 5? 1069
B. 3225
C. 7425
D. 3055
E. 2225
is the range of values for an 8-bit two’s complement integer? 0to255
11010110
A. 10110110 B. 10101010 C. 01010110 D. 00101010 E. 11010110
10. Whatisthefollowingbase3numberinbase7?21013 A. 7367
B. 1237 C. 1217 D. 647 E. 467
11. 6-bittwo’scomplement,signedmagnitude,andunsignedallrepresentthesamenumberofintegers,somejust have more negative than positive.
A. True B. False
CMPE 12 Final Exam – Version A Page 2 of 13 Winter 2019
CruzID:
@ucsc.edu
Data Representation
6. Which IEEE 754 single precision floating point number is furthest from zero?
A. 0xC70FFFFF
B. 0x47700000
C. 0x1F8FFFFF
D. 0x380FFFFF
E. 0xB8700000
A.
B.
C.
D.
E.
Negative edge triggered D flip flop SR latch active high
Positive edge triggered D flip flop Dlatch
SR latch active low
14. Thisfigureislogicallyequivalenttowhichcircuit?
CruzID: @ucsc.edu
Logic Design
12. Thisfigureislogicallyequivalenttowhichcircuit?
A.
B.
C.
D.
E.
XNORgate XORgate ANDgate XORgate Positive D-Latch
13. Whatdevicedoesthistimingdiagramrepresent?
A.
B.
C.
D.
E.
ANDgate
XORgate
Negative D-Flip Flop XNORgate
ORgate
15. Thisfigureislogicallyequivalenttowhichcircuit?
A.
B.
C.
D.
E.
XORgate XORgate Negative D-Latch Positive D-latch XNORgate
CMPE 12 Final Exam – Version A
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CruzID:
@ucsc.edu
16. SelecttheBooleanexpressionmatchingthefilledareasofthisVenndiagram.
A. (Red + Square) · (Big · Red · Square) · (Big + Red + Square)
B. Red·Square·(Big·Red·Square)·(Big+Red+Square)
C. (Red + Square) + (Big · Red · Square) · (Big + Red + Square)
D. Red · Square · (Big · Red · Square)
E. Red·Square·(Big·Red·Square)+(Big+Red+Square)
17. Howmanyoutputsdoesa4-16decoderhave? A. 64
B. 4 C. 1 D. 16 E. 32
Memory
18. How many bits are needed to represent a memory location address in a 4TB memory space that is 64-byte addressable?
A. 28 B. 36 C. 64 D. 34 E. 234
19. Howmuchmemoryisallocatedwiththefollowinglineofcode?
.asciiz “ce_12”
A. 6bytes B. 5words C. 4bytes D. 2words E. 5bytes
CMPE 12 Final Exam – Version A
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A.
B.
C.
D.
E.
Did you ever hear the tragedy of Darth Plagueis the Wise? No! Try not. Do. Or do not. There is no try.
Help me, Obi-Wan Kenobi. You’re my only hope.
I have a bad feeling about this.
I find your lack of faith disturbing.
23. SaythatauserentersasingleASCIIcharacterintherange‘0’-‘9’.Assumethattheuserinputisstoredin$v0. Which MIPS instruction would you use to convert their input into an integer in the range 0-9?
A. subi $t0, $v0, 49 B. subi $t0, $v0, 48 C. addi $t0, $v0, 48 D. subi $t0, $v0, 30 E. subi $t0, $v0, 60
CMPE 12 Final Exam – Version A
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CruzID:
@ucsc.edu
For the following two questions, assume a portion of data memory looks like this:
ADDRESS CONTENTS
0x10011085 0xCD
0x10011084 0xAB
0x10011083 0x87
0x10011082 0x65
0x10011081 0x43
0x10011080 0x21
20. Assumingbigendianmemorystorage,whatisin$t7afterthefollowinginstructions?
ADDI $t0, $zero, 0x10011080
LH $t7, 2($t0)
SW $t7, ($t0)
LW $t7, ($t0)
A. 0x87654321 B. 0x00008765 C. 0x00006587 D. 0xFFFF8765 E. 0x00004321
21. Assuminglittleendianmemorystorage,whatisin$t0afterthefollowinginstructions?
LI $t3, 0x10011082
LW $t0, ($t3)
A. 0x00008765
B. 0x5678BADC
C. 0x6587ABCD
D. Undefined. There will be an alignment error.
E. 0xCDAB8765
ASCII
22. DecodethefollowingASCIIstring.Valuesaregiveninhex.
44 69 64 20 79 6f 75 20 65 76 65 72 20 68 65 61 72 20 74 68 65 20 74 72 61 67 65 64 79 20 6f 66 20 44 61 72 74 68 20 50 6c 61 67 75 65 69 73 20 74 68 65 20 57 69 73 65 3f
A.
ADDIU $1 $0 0xABCD
SRL $1 $1 16
OR $16 $13 $1
B. ORI C. LI
$16 $13 0xABCDEF00
$1 0xABCDEF00
OR $16 $13 $1
D. LUI
OR $16 $13 $1
$1 0xEF00
ORI $16 $16 0xABCD
E. LUI
ORI $1 $1 0xEF00
$1 0xABCD
OR $16 $13 $1
26. Whichregister(s)inMIPSmustthecalleepreserve? A. $t0-$t9
B. $s0-$s7 C. $sp
D. $v0-$v1 E. $a0-$a3
CMPE 12 Final Exam – Version A
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CruzID:
@ucsc.edu
MIPS
24. Whatisthevalueof$t0afterthefollowinginstructionsareexecuted(representedinhex)?
li $t1, 5
li $t0, 5
loop:
sll $t0, $t0, 1
addi $t1, $t1, -1
bgez $t1, loop
li $v0, 10
syscall
A. 0x00000140 B. 0x0001FA00 C. 0x000000A0 D. 0x00001400 E. 0x001000FA
25. WhichMIPS32native/basicinstruction(s)performthesamefunctionasthefollowingpseudoinstruction?
ORI $s0 $t5 0xABCDEF00
CruzID: @ucsc.edu 27. Whatisthevalueof$t0afterthefollowinginstructionsareexecuted?
li $t0, 4
li $t1, 5
add $t0, $t1, $t0
addi $t0, $t0, -1
xor $t0, $t0, $t0
A.
B.
C.
D.
E.
0
6
10
8
Not enough information given
28. Whatistheleastsignificantbytestoredin$t0afterthefollowingMIPScommandsexecute?
li $t0, 0x9F
andi $t0, $t0, 0x0F
A. 00001111 B. 10011111 C. 11110000 D. 00011111 E. 00000000
29. WhatisprintedtothescreenafterthefollowingMIPScommandsexecute?
1 2 3 4 5 6 7 8 9
.data
prompt1: .ascii “I”
prompt2: .asciiz ” LOVE”
prompt3: .asciiz ” CE 12 FINAL”
.text
li $v0, 4
la $a0, prompt1
syscall
A.
B. C. D. E.
ILOVE
I
ILOVE
I LOVE CE12 FINAL nothing
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30. Processinganinstructionrequiresthefollowingsteps
a. Execute operation/evaluate effective address b. Write value to register file
c. Fetch instruction from memory
d. Access data from memory
e. Decode instruction
What is the correct ordering for these steps? A. ecadb
B. ceadb
C. caedb
D. deacb E. aebdc
31. WhichcombinationofMIPSinstructionsperformapopoperationofonewordfromthestack?
A.
B.
C.
sw $t0, ($sp)
subi $sp, $sp, 4
addi $sp, $sp, 4
lw $t0, ($sp)
lw $t0, ($sp)
addi $sp, $sp, 4
D. subi $sp, $sp, 4 sw $t0, ($sp)
E.
The next four questions will refer to the following MIPS code:
1 .text
2 la $a0, str1
3 addiu $v0, $zero, 4
4 syscall
5
6 la $a0, str2
7 syscall
8
9 lbu $a0, str3
10 addiu $v0, $zero, 11
11 syscall
12
13 addiu $v0, $zero, 1
14 syscall
15
16 addiu $v0, $zero, 10
17 syscall
18
19 .data
20 str1: .ascii “hello”
21 str2: .asciiz “there”
22 str3: .byte 0x21 0x21 0x00
none of the above
CMPE 12 Final Exam – Version A
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CruzID: @ucsc.edu
32. Assumeyouchangedline21intheoriginalprogramfrom
str2: .asciiz “there”
to
str2: .ascii “there”
What will be printed to the screen after the altered program completes execution? A. hellothere!!33
B. hellotherethere!33
C. hellothere!33
D. hellothere!!there!!!33 E. hellothere!!there!!33
33. Whatwillbeprintedtothescreenaftertheoriginalprogramcompletesexecution? A. hellotherethere!33
B. hellothere!!33
C. hellotherethere!!33 D. hellothere!33
E. hellotherethere!21
34. Assumeyouchangedline13intheoriginalprogramfrom
addiu $v0, $zero, 1
to
addiu $v0, $zero, 35
What will be printed to the screen after the altered program completes execution? A. hellotherethere!00000000000000000000000000100001
B. hellothere!00000000000000000000000000100001
C. hellothere!!00000000000000000000000000100001
D. hellotherethere!0x00000021 E. hellotherethere!33
35. Giventhebranchinstructioninmachinecode
000101 00010 01000 1111111111111100
Assume the branch target address is 0x2004, what is the address of the branch instruction?
A.
B.
C.
D.
E.
None of the other answers
0x2004
0x2010
0x2018
0x2014
CMPE 12 Final Exam – Version A
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@ucsc.edu
CruzID: @ucsc.edu
The addresses of some of the instructions of the following program are listed. Please refer to the program for the
next two questions
.text
main:
0x00400000 jal getString
move $a0, $v0
li $v0, 4
0x0040000c syscall
li $v0, 10
syscall
0x00400018 getString:
#sets v0 to address of string
la
0x00400020 jr
.data
string1: .asciiz
$v0, string1
$ra
“Greetings!”
36. Whatisthevalueof$pcrightafterthejalistaken? A. 0x0040000c
B. 0x00400000 C. 0x00400018 D. 0x00400020 E. 0x00400004
37. Whatisthevalueof$rarightafterthejalistaken? A. 0x0040000c
B. 0x00400000 C. 0x00400020 D. 0x00400018 E. 0x00400004
CMPE 12 Final Exam – Version A
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A.
B.
C.
D.
E.
9753
’%#!
97531
0x39 0x37 0x35 0x33 0x31 97531/
Instruction Decoding
39. AssumeanISAwith8generalpurposeregistersandthefollowing16-bitinstructionformat:
| opcode | RD | RS | RT |
How many unique instructions can this ISA have? A. 16
B. 9
C. 7
D. 128 E. 8
40. DecodethefollowingMIPS32instruction:0x8D4C3210
A. SW B. AND C. ANDI D. LW E. LW
$t2 0x0101 ($t3)
$t2 0x0123 $t4
$t2 $t4 0x0123
$t4 0x3210 ($t2)
$t2 0x3210 ($t4)
CMPE 12 Final Exam – Version A
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Arrays
38. ThenextquestionreferstothefollowingMIPScode.Assumeallmemorylocationsareinitializedto0x0000.
.text
la $t0, space
li $t1, 0
li $t2, 0x39
loop:
sb $t2, ($t0)
addi $t0, $t0, 1 # increment address
addi $t1, $t1, 1 # incrememt counter
subi $t2, $t2, 2
blt $t1, 5, loop
la $a0, space
li $v0, 4
syscall
li $v0, 10
syscall
.data
space: .space 10
What will be printed to the screen after the program completes execution?
CruzID:
@ucsc.edu
41. DecodethefollowingMIPS32instruction:0x01097820.Selectallthatapply. A. ADD $t0 $t1 $t7
B. AND$8 $9 $15
C. ADD$8 $9 $15
D. ADD $t7 $t0 $t1 E. ADD$15$8 $9
Data Path
42. Assume $t0 = 5 and LB $t0 4($t0) is executed. The programmer has access to all memory locations. What is the value on wire 9?
A.
B.
C.
D.
E.
5
9
Not enough information given 4
8
CMPE 12 Final Exam – Version A
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43. TheinstructionSUBI $t7 $t7 -1isexecuted.Whatisthevalueonwire4?
A.
B.
C.
D.
E.
None of the other answers 0xFFFF
0xF
0xFFFFFFFF
Not enough information given
44. Assume the values on wires 5, 7, 10, 11, and 12 are 0x08, 0x12, 0x1A, 0x1B and 0x1B respectively. Which instruction could correspond to these values?
A.
B.
C.
D. LW $t0 12($t1)
E. LH $t8 8($t9)
45. Assume$s0=0xAB,$s1=0xF4andSW $s1 8($s0)isexecuted.Whatisthevalueonwire8?
A.
B.
C.
D.
E.
0xF4
0x08
Not enough information given 0x10
0xAB
Not enough information given
ADDI $12 $8 18
ADDI $s1 $s2 8
CMPE 12 Final Exam – Version A
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CruzID:
@ucsc.edu
REG NAME REG # MNEMONIC MEANING TYPE OPCODE FUNCT MNEMONIC MEANING TYPE OPCODE FUNCT
$zero 0
sll Logical Shift Left R 0x00 0x00
add Add R 0x00 0x20
$at 1
srl Logical Shift Right (0-extended) R 0x00 0x02
addi Add Immediate I 0x08 NA
$v0 2
sra Arithmetic Shift Right (sign-extended) R 0x00 0x03
addiu Add Unsigned Immediate I 0x09 NA
$v1 3
jr Jump to Address in Register R 0x00 0x08
addu Add Unsigned R 0x00 0x21
$a0 4
mfhi Move from HI Register R 0x00 0x10
and Bitwise AND R 0x00 0x24
$a1 5
mflo Move from LO Register R 0x00 0x12
andi Bitwise AND Immediate I 0x0C NA
$a2 6
mult Multiply R 0x00 0x18
beq Branch if Equal I 0x04 NA
$a3 7
multu Unsigned Multiply R 0x00 0x19
blez Branch if Less Than or Equal to Zero I 0x06 NA
$t0 8
div Divide R 0x00 0x1A
bne Branch if Not Equal I 0x05 NA
$t1 9
divu Unsigned Divide R 0x00 0x1B
div Divide R 0x00 0x1A
$t2 10
add Add R 0x00 0x20
divu Unsigned Divide R 0x00 0x1B
$t3 11
addu Add Unsigned R 0x00 0x21
j Jump to Address J 0x02 NA
$t4 12
sub Subtract R 0x00 0x22
jal Jump and Link J 0x03 NA
$t5 13
subu Unsigned Subtract R 0x00 0x23
jr Jump to Address in Register R 0x00 0x08
$t6 14
and Bitwise AND R 0x00 0x24
lb Load Byte I 0x20 NA
$t7 15
or Bitwise OR R 0x00 0x25
lbu Load Byte Unsigned I 0x24 NA
$s0 16
xor Bitwise XOR (Exclusive-OR) R 0x00 0x26
lh Load Halfword I 0x21 NA
$s1 17
nor Bitwise NOR (NOT-OR) R 0x00 0x27
lhu Load Halfword Unsigned I 0x25 NA
$s2 18
slt Set to 1 if Less Than R 0x00 0x2A
lui Load Upper Immediate I 0x0F NA
$s3 19
sltu Set to 1 if Less Than Unsigned R 0x00 0x2B
lw Load Word I 0x23 NA
$s4 20
j Jump to Address J 0x02 NA
mfc0 Move from Coprocessor 0 R 0x10 NA
$s5 21
jal Jump and Link J 0x03 NA
mfhi Move from HI Register R 0x00 0x10
$s6 22
beq Branch if Equal I 0x04 NA
mflo Move from LO Register R 0x00 0x12
$s7 23
bne Branch if Not Equal I 0x05 NA
mult Multiply R 0x00 0x18
$t8 24
blez Branch if Less Than or Equal to Zero I 0x06 NA
multu Unsigned Multiply R 0x00 0x19
$t9 25
addi Add Immediate I 0x08 NA
nor Bitwise NOR (NOT-OR) R 0x00 0x27
$k0 26
addiu Add Unsigned Immediate I 0x09 NA
or Bitwise OR R 0x00 0x25
$k1 27
slti Set to 1 if Less Than Immediate I 0x0A NA
ori Bitwise OR Immediate I 0x0D NA
$gp 28
sltiu Set to 1 if Less Than Unsigned ImmediateI 0x0B NA
sb Store Byte I 0x28 NA
$sp 29
andi Bitwise AND Immediate I 0x0C NA
sh Store Halfword I 0x29 NA
ori Bitwise OR Immediate I 0x0D NA
sll Logical Shift Left R 0x00 0x00
xori Bitwise XOR (Exclusive-OR) Immediate I 0x0E NA
slt Set to 1 if Less Than R 0x00 0x2A
lui Load Upper Immediate I 0x0F NA
slti Set to 1 if Less Than Immediate I 0x0A NA
mfc0 Move from Coprocessor 0 R 0x10 NA
sltiu Set to 1 if Less Than Unsigned ImmediateI 0x0B NA
lb Load Byte I 0x20 NA
sltu Set to 1 if Less Than Unsigned R 0x00 0x2B
lh Load Halfword I 0x21 NA
sra Arithmetic Shift Right (sign-extended) R 0x00 0x03
lw Load Word I 0x23 NA
srl Logical Shift Right (0-extended) R 0x00 0x02
lbu Load Byte Unsigned I 0x24 NA
sub Subtract R 0x00 0x22
lhu Load Halfword Unsigned I 0x25 NA
subu Unsigned Subtract R 0x00 0x23
sb Store Byte I 0x28 NA
sw Store Word I 0x2B NA
sh Store Halfword I 0x29 NA
xor Bitwise XOR (Exclusive-OR) R 0x00 0x26
sw Store Word I 0x2B NA
xori Bitwise XOR (Exclusive-OR) Immediate I 0x0E NA
R Type: instr rd rs rt (arithmetic, logical)
instr rd rt shamt (shifts)
31 26 25 21 20 16 15 11 10 6 5 0 <- 6 bits -> <- 5 bits -> <- 5 bits -> <- 5 bits -> <- 5 bits -> <- 6 bits ->
opcode
rs
rt
rd
shamt
funct
I Type: instr rt rs immediate
branch rs rt immediate
instr rt immediate(rs)
(arithmetic, logical)
(branches)
(loads, stores)
31 26 25 21 20 16 15 0
<- 6 bits ->
<- 5 bits ->
<- 5 bits ->
<- 16 bits ->
opcode
rs
rt
immediate
J Type: j immediate (jumps)
31 26 25 0 <- 6 bits -> <- 26 bits ->
opcode
immediate
ASCII CODE ASCII CODE ASCII CODE – ASCII CODE
BIN OCT DEC HEX CHARACTER BIN OCT DEC HEX CHARACTER BIN OCT DEC HEX CHARACTER BIN OCT DEC HEX CHARACTER
010 0000 40 32 20 space
011 1000 70 56 38 8
101 0000 120 80 50 P
110 1000 150 104 68 h
010 0001 41 33 21 !
011 1001 71 57 39 9
101 0001 121 81 51 Q
110 1001 151 105 69 i
010 0010 42 34 22 ”
011 1010 72 58 3A :
101 0010 122 82 52 R
110 1010 152 106 6A j
010 0011 43 35 23 #
011 1011 73 59 3B ;
101 0011 123 83 53 S
110 1011 153 107 6B k
010 0100 44 36 24 $
011 1100 74 60 3C <
101 0100 124 84 54 T
110 1100 154 108 6C l
010 0101 45 37 25 %
011 1101 75 61 3D =
101 0101 125 85 55 U
110 1101 155 109 6D m
010 0110 46 38 26 &
011 1110 76 62 3E >
101 0110 126 86 56 V
110 1110 156 110 6E n
010 0111 47 39 27 ‘
011 1111 77 63 3F ?
101 0111 127 87 57 W
110 1111 157 111 6F o
010 1000 50 40 28 (
100 0000 100 64 40 @
101 1000 130 88 58 X
111 0000 160 112 70 p
010 1001 51 41 29 )
100 0001 101 65 41 A
101 1001 131 89 59 Y
111 0001 161 113 71 q
010 1010 52 42 2A *
100 0010 102 66 42 B
101 1010 132 90 5A Z
111 0010 162 114 72 r
010 1011 53 43 2B +
100 0011 103 67 43 C
101 1011 133 91 5B [
111 0011 163 115 73 s
010 1100 54 44 2C ,
100 0100 104 68 44 D
101 1100 134 92 5C \
111 0100 164 116 74 t
010 1101 55 45 2D –
100 0101 105 69 45 E
101 1101 135 93 5D ]
111 0101 165 117 75 u
010 1110 56 46 2E .
100 0110 106 70 46 F
101 1110 136 94 5E ^
111 0110 166 118 76 v
010 1111 57 47 2F /
100 0111 107 71 47 G
101 1111 137 95 5F _
111 0111 167 119 77 w
011 0000 60 48 30 0
100 1000 110 72 48 H
110 0000 140 96 60 `
111 1000 170 120 78 x
011 0001 61 49 31 1
100 1001 111 73 49 I
110 0001 141 97 61 a
111 1001 171 121 79 y
011 0010 62 50 32 2
100 1010 112 74 4A J
110 0010 142 98 62 b
111 1010 172 122 7A z
011 0011 63 51 33 3
100 1011 113 75 4B K
110 0011 143 99 63 c
111 1011 173 123 7B {
011 0100 64 52 34 4
100 1100 114 76 4C L
110 0100 144 100 64 d
111 1100 174 124 7C |
011 0101 65 53 35 5
100 1101 115 77 4D M
110 0101 145 101 65 e
111 1101 175 125 7D }
011 0110 66 54 36 6
100 1110 116 78 4E N
110 0110 146 102 66 f
111 1110 178 126 7E ~
011 0111 67 55 37 7
100 1111 117 79 4F O
110 0111 147 103 67 g
111 1111 177 127 7F DEL
CODE IN
SERVICE $v0 ARGUMENTS RESULT
print integer 1 $a0 = integer to print
print float 2 $f12 = float to print
print double 3 $f12 = double to print
$a0 = address of null-terminated string to
print string 4 print
read integer 5 $v0 contains integer read
read float 6 $f0 contains float read
read double 7 $f0 contains double read
$a0 = address of input buffer
read string 8 $a1 = maximum number of characters to read See note below table
sbrk
(allocate $v0 contains address of allocated heap memory) 9 $a0 = number of bytes to allocate memory
exit
(terminate execution) 10
print
character 11 $a0 = character to print See note below table
read
character 12 $v0 contains character read
$a0 = address of null-terminated string
containing filename
$a1 = flags $v0 contains file descriptor (negative
open file 13 $a2 = mode if error). See note below table
$a0 = file descriptor
read from $a1 = address of input buffer
file 14 $a2 = maximum number of characters to read
$v0 contains number of characters read
(0 if end-of-file, negative if error).
See note below table
$a0 = file descriptor
$a1 = address of output buffer write to file 15 $a2 = number of characters to write
$v0 contains number of characters
written (negative if error). See note
below table
close file 16 $a0 = file descriptor
exit2
(terminate
with value) 17 $a0 = termination result See note below table
Services 1 through 17 are compatible with the SPIM simulator, other than Open File (13) as described in the Notes below the table. Services 30 and higher are exclusive to MARS.
$a0 = low order 32 bits of system time time (system $a1 = high order 32 bits of system
time) 30 time. See note below table
$a0 = pitch (0-127)
$a1 = duration in milliseconds
$a2 = instrument (0-127) Generate tone and return immediately.
MIDI out 31 $a3 = volume (0-127) See note below table
Causes the MARS Java thread to sleep
for (at least) the specified number of
milliseconds. This timing will not be
$a0 = the length of time to sleep in precise, as the Java implementation sleep 32 milliseconds. will add some overhead.
$a0 = pitch (0-127)
$a1 = duration in milliseconds
MIDI out $a2 = instrument (0-127) Generate tone and return upon tone synchronous 33 $a3 = volume (0-127) completion. See note below table
print integer
in
hexadecimal 34
$a0 = integer to print
Displayed value is 8 hexadecimal
digits, left-padding with zeroes if
necessary.
print integer Displayed value is 32 bits, left- in binary 35 $a0 = integer to print padding with zeroes if necessary.
print integer
as unsigned 36 $a0 = integer to print Displayed as unsigned decimal value.
MIPS Address Space (Not to Scale)
0xffff ffff
0xffff 0000
0x9000 0000
0x7fff fe00
0x1004 0000
0x1001 0000
MMIO
Kernel Data
Stack
Heap
Static Data
Program Text
Reserved
0x0040 0000
0x0000 0000