1. CS334: Memory _ Mars simulator Lab 4(part2-2)

2. Control Structures in mips. Load / Store Instructions Add in hex.
Contents:
Control Structures in mips.
Load / Store Instructions
Add in hex.
Part2:memory.
Exercise.
Questions.
Ins.Ebtesam AL-Etowi

3. Control Structures Branches
comparison for conditional branches is built into instruction.
Jump to lable(target) if condition true else continue
Ins.Ebtesam AL-Etowi

4. Cont……. Jumps j target # unconditional jump to program label target
Ins.Ebtesam AL-Etowi

5. Ins.Ebtesam AL-Etowi

6. Load / Store Instructions
RAM access only allowed with load and store instructions .
all other instructions use register operands
load:
lw register_destination, RAM_source
#copy word (4 bytes) at source RAM location to destination register.
store word:
sw register_source, RAM_destination
#store word in source register into RAM destination
Ins.Ebtesam AL-Etowi

7. Cont ….. load immediate li register_destination, value example:
#load immediate value into destination register
example:
lw $t0, var1 # load contents of RAM location into register $t0: $t0 = var1
li $t1, # $t1 = 5 ("load immediate")
sw $t1, var1 # store contents of register $t1 into RAM: var1 = $t1
Ins.Ebtesam AL-Etowi

8. Add hex digit 3A49 + 4BA9 -------- 85F2
Before working with memory, you need to be able to add hex digits.
Here is an example of adding 2 hex digits:
3A49
+ 4BA9
85F2
Ins.Ebtesam AL-Etowi

9. Add hex digit 8* 16^3 + 5 * 16^2 + F * 16^1 + 2* 16^0 = 34290
Note that numbers without the 0x are decimal:
Digit 0: 0x9 + 0x9 = = 18 = 0x12 (1 * 16^1 + 2 * 16^0). So, write down the 2 and carry the 1.
Digit 1: 1 (carry) + 0x4 + 0xA = = 15 = 0xF. So, write down the F. There is no carry.
Digit 2: 0xA + 0xB = = 21 = 0x15 (1*16^1 + 5 * 16^0). So, write down the 5 and carry the 1.
Digit 3: 1 (carry) + 0x3 + 0x4 = = 8 = 0x8. So, write down the 8.
8* 16^3 + 5 * 16^2 + F * 16^1 + 2* 16^0 = 34290
34290=0x 85F2
Ins.Ebtesam AL-Etowi

10. Ins.Ebtesam AL-Etowi

11. Part 2: memory
Now, let's look at memory. Start by entering the following into the simulator:
.data
A: .word 25, 3, 0x44, 0x33, 0x25, 0x3, 16, 0x22, 44, 33, 22
Ins.Ebtesam AL-Etowi

12. Ins.Ebtesam AL-Etowi Words
Hex Value (show the correct number of digits! 8 hex digits = 32 bits = 1 word) 25 3
0x44
0x33
0x25
0x3 16
0x22 44 33 22
Ins.Ebtesam AL-Etowi

13. Cont….
".data" is an assembler directive (instruction to the assembler) that tells the assembler we are in the data segment of memory. ".word" is an assembler directive that tells the assembler to store the following into subsequent words in memory. The 0x values are in hex. The other values are in decimal. The data segment of memory begins at address 0x
Ins.Ebtesam AL-Etowi

14. Cont..)) Questions
Question 8:Now that you understand what you are looking at, fill out the above table (if you have already filled out that table, then you are done!)
(show the correct number of digits! 8 hex digits = 32 bits = 1 word)
Ins.Ebtesam AL-Etowi

15. Cont…..
25 0x
3 0x
0x44 0x
0x33 0x
0x25 0x
0x3 0x
16 0x
0x22 0x
44 0x C
33 0x
22 0x
Ins.Ebtesam AL-Etowi

16. Questions
Question 9:
2A8E
+ CBF3
????
Answer: F681
Question 10: : Does each box shown in the data segment window represent a byte or a word? Please
explain.
Answer: A word. It packs four bytes into a word and shows it on each of the boxes.
Ins.Ebtesam AL-Etowi

17. exercise to perform the following: If (n1>n2) n2= 0x5 true Else
Add it n1+n2
true
in .data
n1 =8
n2 =11
Ins.Ebtesam AL-Etowi

18. Exercise if the condition false
Ins.Ebtesam AL-Etowi

19. Execute
Ins.Ebtesam AL-Etowi

20. Exercise if the condition true
Ins.Ebtesam AL-Etowi

21. Ins.Ebtesam AL-Etowi

22. ? ? ?
Ins.Ebtesam AL-Etowi