HW2

Q1)

A)

from the green card:

code	Width (bits)	Binary
i) opcode (lw)	7	0000011
ii) rd (x14)	5	01110
iii) rs1 (x10)	5	01010
iv) imm[11:0] (16)	12	000000010000
v) we know its in the format `imm[11:0]	rs1	funct3

• 00000001000001010010011100000011

B)

converting from binary to hexadecimal and grouping we get 0x01052703

C)

from part b we get the hex then we get each digit and multiply it by $16^{n-1}$ to get the unsigned decimal

• $0x01052703=0\times 16^7+1\times 16^6+0\times 16^5+5\times 16^4+2\times 16^3+7\times 16^2+0\times 16^1+3\times 16^0$
  • Unsigned decimal = 17 114 883

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Q2)

A)

im going to use the method of dividing by 2 then taking the remainder as the bit and since we are dividing by 2 we read from bottom as msb and top as lsb

Step	Quotient	Remainder (bit)
28 215 603 ÷ 2	14 107 801	1
14 107 801 ÷ 2	7 053 900	1
7 053 900 ÷ 2	3 526 950	0
3 526 950 ÷ 2	1 763 475	0
1 763 475 ÷ 2	881 737	1
881 737 ÷ 2	440 868	1
440 868 ÷ 2	220 434	0
220 434 ÷ 2	110 217	0
110 217 ÷ 2	55 108	1
55 108 ÷ 2	27 554	0
27 554 ÷ 2	13 777	0
13 777 ÷ 2	6 888	1
6 888 ÷ 2	3 444	0
3 444 ÷ 2	1 722	0
1 722 ÷ 2	861	0
861 ÷ 2	430	1
430 ÷ 2	215	0
215 ÷ 2	107	1
107 ÷ 2	53	1
53 ÷ 2	26	1
26 ÷ 2	13	0
13 ÷ 2	6	1
6 ÷ 2	3	0
3 ÷ 2	1	1
1 ÷ 2	0	1
then reading from msb to lsb we get `00000001 10101110 10001001 00110011
`

B)

for this we can group them into 4 nibbles :) then get their corresponding hex value by knowing each 4 bits giving 16 in length

Binary	Hex
0000	0
0001	1
1010	A
1110	E
1000	8
1001	9
0011	3
0011	3
then adding the 0x telling it the numbers are hex we get 0x01AE8933

C)

i)

• to find the operation we can look at the last 7 bits first to determine the opcode as its the same for all types of instructions, and then first 7 to determine if its a funct7 or an immediate
  • 00000001 10101110 10001001 00110011 and looking at this we can see the last 7 being indicative of an r type so we check the first 7 bits for operation, which matches add

ii)

• to find the destination its the rd, and that is the 11th bit to the 7th bit which is 10010 which gives us x18 register

iii)

• to find the 2 sources we can find using the bits for rs1 19-15 and rs2 24-20 and converting that to decimal gives us x29, x26

iv)

• thus the full operation is
  • add x18, x29, x26

3-4)

• i would love to do a full write up for 3-4 but i just dont have time, and since they are not marked im not going to post them sorry </3 if they are supposed to be marked but as a mistake or smth
• 
  • here is my proof if you would like it ill do it :) because practice is good