MIPS

Question 1

How long are MIPS addresses

Answer 1

32 bits

Question 2

how long are mips instructions?

Answer 2

32 bits

Question 3

How is a MIPS instruction held in memory?

Answer 3

Held in 4 consecutive memory locations

Question 4

How many registers inside a mips?

Answer 4

32($0-$31)

Question 5

What can arithmetic operations operate on?

Answer 5

Data read from registers

Question 6

What is RISC based architecture?

Answer 6

“load and store”, only load and store instructions access memory, all other instructions access registers

Question 7

Describe R-type instructions

Answer 7

Register operands - all arithmetic
op = operation or type of the instruction
rs = first register argument (source)
rt = second register argument (source)
rd = result of the operation (destination)
shamt = shift amount (ignore at present)
funct = particular type of the operation

op always equal to zero, so funct describes what the instruction does

Question 8

Name the key difference between add and addu

Answer 8

addu stands for add unsigned, add is assumed to be working with 2’s complement singed numbers

Real difference is that signed add “Throws interrupt” whenever overflow occurs for the signed operation

addu ignores the overflow

Question 9

How do you calculate the memory address?

Answer 9

sum the displacement value given and the value stored inside the register.

Question 10

If the displace read 0x0008 how many words is that

Answer 10

2 words displacement

Question 11

What are I type instructions used for?

Answer 11

memory access instructions
instructions that require a constant value
branch instructions

Question 12

consider I type
lw $20, 0x1234($5)
in what order will the op, rs, rt, and offset be

Answer 12

lw, 5, 20, 0x1234

Question 13

Describe how lw instruction

Answer 13

lw $10, 0($x)
– copies the word from address Z into register $10.
– A high-level language view would be that array element Z[0] is
copied into register 10.

Question 14

Describe how sw instruction

Answer 14

sw $10, 80($x)
– copies the value from register $10 into the memory location 20
word offsets from Z.
– A high-level language view would be that the contents of register
10 is copied into array element Z[20].

Question 15

What is an I type (immediate value)?

Answer 15

instead of off set, actual value is passed into instruction to be added or whatever to other register and store in register

Question 16

What is the offset when concerning a branch command?

Answer 16

the NUMBER of INSTRUCTIONS to branch over starting from the instruction following the branch
PC = PC+ 4 + 0xFFFFABCD * 4

Question 17

How do we store a 32 bit address in a register?

Answer 17

Use a load upper instruction and then use OR instruction with other half of immediate and store in the same register

Question 18

At what address is the executable code loaded from?

Answer 18

0x00400000, the text segment

Question 19

What is an assembly directive?

Answer 19

They tell the assembler what to do

Question 20

What are the two key directives

Answer 20

.data, .text, they are concerned with the way the assembly language program is laid out in memory

Question 21

Where is the data loaded at?

Answer 21

0x1001000, known as the dynamic data segment

Question 22

What is the .word directive do?

Answer 22

Its a data layout directive, it allows the assembly programmer to describe the data in a more natural way then binary,
in this case, the 32 bit words that follow are written into the dynamic data segment of the program
vals: .word 4, 5, 8, 0

Question 23

describe what the assembly directive .word 10,20,30 do?

Answer 23

Store 10,20,30 in successive memory locations as 32 bit words

Question 24

What is an assembly label?

Answer 24

A colon after a word indicates an address label, this is a symbolic representation of that address.
These can be used withing the assembly code and the assembler will put in the correct address values

Question 25

What does the assembly directive .globl do?

Answer 25

Tells the assembly that this is a global symbol, one which needs to accessed outside of the current assembly program, e.g. .globl main

Question 26

Describe the J-Type instruction

Answer 26

Consist of a 6 bit op and a 26 bit target, used to jump

Question 27

explain what the instruction j target does

Answer 27

jumps to target, multiplies the 26bit target by 4 and stores it in the 28 LSB in the program counter

Question 28

explain what the jal target instruction does

Answer 28

Jump and link target, stores the return address in the ra register, ie PC + 4 and then changes the program counter to the 28 bit target

Question 29

Describe a procedure being called using the jal and jr instructions

Answer 29

the jal instruction stores the return address of the line of code currently after the branch, i.e pc +4 .
The PC is then changed to the operand provided, after the procedure has run the code it needs to, the jr $ra instruction is called which stores the ra inside the PC