Week 11: LDR/STR encoding/decoding, Examples, and Stacks

Question 1

For arm LDR / STR instructions, whenever a single byte is loaded into a register, the most significant bits are…

Answer 1

Set to zero

Question 2

For ARM LDR / STR instructions, are register-based dynamic shifts allowed?

Answer 2

NO

Question 3

For ARM LDR / STR instructions, are static shifts allowed?

Answer 3

YES

Question 4

For ARM LDR / STR instructions, when P = 0 (post-indexed addressing), what is true?

Answer 4

The W (write-back) bit is ALWAYS set to zero

Question 5

When will the W-bit always be 0?

Answer 5

When P=0

Question 6

For ARM LDR / STR instructions, an immediate-offset is always

Answer 6

A 12-bit literal offset, there is no rotation

Question 7

To calculate the absolute value in ARM, what can you do?

Answer 7

TEQ r0, #0
RSBMI r0, r0, #0

Question 8

To clear higher-order bytes, what operation can you use?

Answer 8

AND

AND r0,r0,#0xFF
AND r1,r1,#0xFF

Question 9

To clear specific bytes, what can you use?

Answer 9

BIC

BIC r2,r2,#0xFF0000
BIC r2,r2,#0xFF000000

Question 10

Is the following valid?

BIC r2,r2,#0xFFFF0000

Answer 10

No, because only instructions that are 16-bits long can be used (FF)

Question 11

To swap variables without using an intermediate register, what can be done?

Answer 11

ADD r0, r0, r1

SUB r1, r0, r1
SUB r0, r0, r1

Question 12

To multiply by 2^n -1, 2^n, or 2^n + 1

What can be done?

Answer 12

For 2^n
Simply use a mov and LSL
MOV r2, r1, LSL #n

For 2^n + 1, use an ADD instruction
ADD r2, r1, r1, LSL#n

For 2^n - 1 use RSB
RSB r2, r1, r1, LSL#n

Question 13

How do you divide by D?

Answer 13

Use MUL and ADR instructions:

[r0] / D = [r0] × (2^N/D) / 2^N

Select N to be a large integer at the same time not to cause an overflow when
evaluating [r0] × (2^N/D)

Evaluate [r0] × (2^N/D)

Arithmetic shift right the result N time

Question 14

How do you convert a capital letter to a small letter?

If the character to be converted to a smaller letter is in r0 and r1 is a working register?

Answer 14

CMP r0, #’A’

RSBGES r1, r0, #’Z’

ORRGE r0, r0, #2_100000

First test if its greater than A

Then test if its between A-Z

If so, force bit 5 to 1 to make it lower case

Question 15

How would you write out the following code in one line?

if (x < 0) x = 0;

Answer 15

BIC r0, r0, r0, ASR#31

The ASR#31 will fill the entire word with only the sign bit

If positive, the result will be 0x00000000

IF negative the result will be 0xFFFFFFFF

Question 16

How do you implement a switch statement in ARM?

Answer 16

Question 17

The stack is a ____ queue, which means items enter through ______ and leave through ____

Answer 17

The stack is a LIFO queue, which means items enter through one end and leave through the same end

Question 18

Stacks are implemented using the

Answer 18

Stack Pointer

Question 19

The push and pop of a stack that grows up and already occupies memory looks like what?

Answer 19

PUSH

STR R0, [SP, #-4]!

POP

LDR R0, [SP], #4

Question 20

The push and pop of a stack that grows up and does not occupy memory looks like what?

Answer 20

PUSH

STR R0, [SP], #-4

POP

LDR R0, [SP, #4]!

Question 21

The push and pop of a stack that grows down and already occupies memory looks like what?

Answer 21

PUSH

STR R0, [SP, #4]!

POP

LDR R0, [SP], #-4

Question 22

The push and pop of a stack that grows down and does not already occupy memory looks like what?

Answer 22

PUSH

STR R0, [SP], #4

POP

LDR R0, [SP, #-4]!

Question 23

The two decisions that must be made when implementing a stack are:

Answer 23

Whether the stack is going to grow up or down

Whether or not the stack pointer is going to point to empty memory or full memory

Question 24

What is the difference between CISC and RISC processors when it comes to the stack?

Answer 24

CISC processors automatically maintain the stack.

RISC processors force the programmer to maintain the stack.