lecture 8

Question 1

3 compilation steps

Answer 1

1. compiler
2. assembler
3. linker

Question 2

what enters the compiler?

Answer 2

filename.c –> your program in high level language (text)

Question 3

what is the gcc compiler command?

Answer 3

gcc -s filename.c

Question 4

what does the gcc -s filename.c create?

Answer 4

when this is entered into the terminal and completes, an assembly program is created (filename.c –> filename.s)

Question 5

what does the compiler change your program to?

Answer 5

filename.s –> assembler program

Question 6

what is the .s/assembler program?

Answer 6

set of text instructions used to program the processor

Question 7

what enters the assembler?

Answer 7

filename.s (assembly program)

Question 8

what is the gcc command for the assembler?

Answer 8

gcc -c filename.s

Question 9

what does the gcc -c filename.s create?

Answer 9

filename.o –> object program/machine program (binary)

Question 10

what is the .o/object program?

Answer 10

machine program –> set of binary instructions that configure and control hardware

Question 11

what enters the linker?

Answer 11

filename.o –> machine program (binary)

Question 12

what is the gcc command for the linker?

Answer 12

gcc filename.o

Question 13

what does the gcc filename.o create?

Answer 13

executable program (a.out)

Question 14

who is the intended user for the machine program?

Answer 14

hardware (computer)

Question 15

highest level language

Answer 15

highest level of abstraction, closest to human language, java python etc
your code basically
text

Question 16

assembly language

Answer 16

lowest level of abstraction, specific to processor architecture (RISC CISC), syntax is human readable but language of machine
text

Question 17

machine language

Answer 17

no abstraction, not human readable, syntax is binary encoded instructions and data, configure and control hardware (MIPS, processor, specific)
binary

Question 18

instruction set architecture

Answer 18

combines instructions with registers, addressing modes, data types

Question 19

MIPS

Answer 19

microprocessor without interlocked pipelined stages
RISC architecture

Question 20

3 MIPS Instruction Types:

Answer 20

1. R-TYPE
2. I-TYPE
3. J-TYPE

Question 21

what is R-TYPE?

Answer 21

instruction operands are registers

Question 22

what is I-TYPE?

Answer 22

instruction operands are registers and a constant (immediate) value

Question 23

anatomy of R-TYPE instruction

Answer 23

operation $1, $2, $3

$1 –> destination operand
$2 and $3 –> source operands
(variables, arguments, etc)

Question 24

anatomy of I-TYPE instruction

Answer 24

operation i $1, $2, 1

i –> immediate (how to tell its I-TYPE)
$1 –> destination operands
$2 –> source operands
1 –> immediate operand (constant)

Question 25

what does the instruction mean?

Answer 25

primitive operation - specify operation and its operands --> operands = necessary variables to perform the operation

Question 26

what are operands?

Answer 26

necessary variables to perform the operation

Question 27

types of operands

Answer 27

immediate (data), source and destination (registers)

Question 28

what is the register symbol

Answer 28

$

Question 29

how many registers does MIPS have?

Answer 29

32 general purpose registers

Question 30

what does the assembler do with the MIPS instructions?

Answer 30

assembler translates MIPS to machine language

Question 31

what step in the compilation system translates code to MIPS?

Answer 31

compiler --> filename.s = MIPS instructions

Question 32

c program: int a = 10; int b = 2; int c = a + b + 2; MIPS Program: addi $8, $0, 10 ^ what does this line mean?

Answer 32

addi $8, $0, 10 $8 --> register $8 holds the value in variable a (a = 10) register $8 = $0(0) + 10 = 10 (AKA a)

Question 33

c program: int a = 10; int b = 2; int c = a + b + 2; MIPS Program: addi $8, $0, 10 addi $9, $0, 2 ^ what does this line mean?

Answer 33

addi $9, $0, 2 $9 --> register $9 holds the value in variable b (b = 2) register $9 = $0(0) + 2 = 2 (AKA b)

Question 34

c program: int a = 10; int b = 2; int c = a + b + 2; MIPS Program: addi $8, $0, 10 addi $9, $0, 2 add $10, $8, $9 ^ what does this line mean?

Answer 34

add $10, $8, $9 register $10 --> holds value in variable c (int c = a + b) $10 = $8 + $9 c = a + b = 10 + 2

Question 35

c program: int a = 10; int b = 2; int c = a + b + 2; MIPS Program: addi $8, $0, 10 addi $9, $0, 2 add $10, $8, $9 addi $10, $10, 2 ^what does this line mean?

Answer 35

addi $10, $10, 2 register $10 --> holds value in variable c (a + b) from line above *now adding 2 to this register since its c = a + b + 2 (cant do this in one MIPS instruction --> had to split it into 2) $10 = $10 + 2 c = (a + b) + 2 = (10 + 2) + 2

Question 36

what is always the opcode for R-TYPE?

Answer 36

000000

Question 37

what are the parts of a 32-bit R-TYPE instruction?

Answer 37

opcode (6) rs (5) rt (5) rd (5) shamt (5) func (6)

Question 38

what are the parts of a 32-bit I-TYPE instruction?

Answer 38

opcode (6) rs (5) rt (5) immediate (16)

Question 39

once the program is in MIPS (assemble program), what does the assembler then do?

Answer 39

translates MIPS to machine language (binary)

Question 40

in the MIPS cheat sheet, what does each part in R-type and I-type get translated to?

Answer 40

its binary equivalent ex) $8 = 01000 for rt (5)

Question 41

what is rt for I-Type?

Answer 41

destination register

Question 42

what is rd for R-type?

Answer 42

destination register

Question 43

what is rs for both R-type and I-type?

Answer 43

source register

Question 44

what is immediate for I-type?

Answer 44

constant

Question 45

what is func for R-Type?

Answer 45

operation (add)

Question 46

what is shamt for R-type?

Answer 46

amount of shift, set to zero in all non-shift instructions