I'm working on this disassembling assignment and a bit confused.
Disassemble the following machine code into operations and arguments,
e.g., ADD R1,R2,R3.Explain what the whole program does,either in plain
English or in C type of pseudo code Note: you need to make up unique
labels for your assembly code.
Add r: Code
0x00: 07 FF
0x02: 06 08
0x04: 28 08
0x06: 38 02
0x08: 02 47
0x0A: 3C 00
here's the simple processor
w - Write back ALU output to register file
src1 - Address for first ALU operand
src2 - Address for second ALU operand
dst - Address in where output is written
Here are the opcodes used in the assignment
ADD 0x0 R[src1] + R[src2]->R[dst]
SUB 0x1 R[src1] – R[src2]->R[dst]
BLEZ 0xA If R[src1] ≤ 0, branch to BAddr (The last 6 bits are the
address)
HALT 0xF
Here's what I got
0000 0111 1111 1111
0000 0110 0000 1000
0001 1000 0000 0008
0011 1000 0000 0001
0000 0001 0010 0111
0011 1110 0000 0000
I get confused here - 0000 0111 1111 1111 looking at the instruction table, it looks like the program is subtracting from the same register and placing the result in the same register. Am I right? I feel like it is wrong. Thanks in advance!
I did not check your opcodes but what you are describing is one way of reseting register to zero. Which makes sense as you do not know what that register contains. Other times XOR operation is used to get the same result.
Related
This is a question and answer from my operating system's textbook:
Question:
A certain computer provides its users with a virtual-memory space of 232 bytes.
The computer has 218 bytes of physical memory. The virtual memory is implemented by paging,
and the page size is 4096 bytes. A user process generates the virtual address 11123456.
Explain how the system establishes the corresponding physical location.
Distinguish between software and hardware operations.
Answer:
The virtual address in binary form is
0001 0001 0001 0010 0011 0100 0101 0110
When I plug "11123456" into a decimal to binary converter (like here: https://www.rapidtables.com/convert/number/decimal-to-binary.html), this is the result: 101010011011101100000000
which differs from the book answer.
Also, when I manually convert using a method like here: https://indepth.dev/the-simple-math-behind-decimal-binary-conversion-algorithms/
I still get something different.
I'm just confused how that virtual address was converted into binary form......
Thanks!
You think
11123456
is a decimal number but it isn't. The Computer representation of number is ever binary, octal or hexadecimal. In this case is Hexadecimal. If you convert the above number from hexadecimal to binary form you obtain the result:
0001 0001 0001 0010 0011 0100 0101 0110
Explanation:
Hexadecimal -> Binary
1 -> 0001
2 -> 0010
3 -> 0011
4 -> 0100
5 -> 0101
6 -> 0110
...
I have been given the hexadecimal number 0xAA and been told to perform the following operations on it, in order:
Shift right by 3
And with 0x18
Shift left by 2
Here are my steps for doing this:
1.) Represent 0xAA in binary
0xAA = 1010 1010
2.) Shift right by 3
0001 0101
3.) Represent 0x18 in binary
0x18 = 0001 1000
4.) Perform 0001 0101 and 0001 1000
0001 0101 & 0001 1000 = 0001 0000
5.) Shift left by 2
0100 0000
And that is my final answer, 0100 0000. However, the answer I am told I should get is 0111. I cannot figure out where I am going wrong, or how I should get that answer. I was hoping someone could tell me where I am going wrong. Thank you all very much.
It seems that I am calculating the correct result, and it is the expected result that is incorrect. Thank you all for your help.
I am in a CSCI class and we are just learning about program execution. I am running a program called "Brookshear machine simulator" which was written by the author of the class text book ( Computer Science 11th edition by J. Glenn Brookshear). The program is intended to add the contents of 11 and 0F, storing the result into F1. I have done everything necessary and produced the hex value in 11 which is 09. I am then asked to convert this into two's complement 8-bit binary, which is where I am having a problem. I will be needing to convert some hex values into two's compl 8-bit binary in the future for this lab but I cant figure out how to do it. Can someone please help me understand what twos comp is and how is it related or the same as 8-bit binary , so I can convert this to two's complement 8-bit binary?
Here is a picture of the machine simulator with the inputs as directed by the lab instructions. My task is to find the hex value in 11 (09) then convert it to twos complement 8-bit binary.
Each hexadecimal digit has a 4 bit binary equivalent:
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
A 1010
B 1011
C 1100
D 1101
E 1110
F 1111
So if you have a two character hex value, like 09 then you can see that 0 = 0000 and 9 = 1001, so that would be:
00001001
which is an 8 bit value.
This works for any length of hex number of course, so for example 37FF in hex would be 0011011111111111 in binary.
Note that two's complement is irrelevant for your example as the number is positive.
I'm not sure if I can ask a binary question here but here goes..
We had this question on our midterm but our professor hasn't provided a correct answer for it. It's been driving me crazy and the final is coming soon so it might be a good idea to fill this gap. Thanks!
Find the smallest two's complement number that, when added to 0101 0101 would result in an overflow. Express your answer in binary.
My reasoning:
I found the range of the original binary 0101 0101 by converting it to an actual number and then added one to it. Then I converted the number that was 1 more than the range into 8-bit binary as my answer. However, this only earned me 3/6 marks. I have no idea what else I could've done. Any insights would be greatly appreciated!
The original binary is a positive number (0 sign bit). Overflow occurs when you add a positive number to it that changes the sign bit. It should be easy to see what the smallest number is using binary notation:
No overflow:
0101 0101
+ 0010 1010
---------
0111 1111
Overflow:
0101 0101
+ 0010 1011
---------
1000 0000
I have no idea if this is what your prof was looking for. (You can probably just subtract from 1000 0000 instead of looking at it as a pattern.)
EDIT Since you asked for an example (meaning something different from the above), here's how subtraction would work:
1000 0000 (the target overflow quantity)
- 0101 0101 (the original binary)
---------
0010 1011 (the smallest number that will overflow when added to original)
That number is 85 in decimal, so 128-85 is 43
I've been looking for the answer quite a while and finally I decided to ask and see if someone here can help me ;)
The format for the conditional jump instruction in MIPS, lets say a beq instruction, contains a 16-bit field that indicates de "size" of the jump. I know it must be added to the current PC in order to obtain the desired direction. My doubt is, how do I set the field for a backwards jump? I guess two's complement? I found this example, but it confused me more than I was... if anyone could explain it briefly I'd be very thankful! By the way, sorry about my grammar/expressions, I'm not very good at english yet.
The example mentioned:
PC+Jump(Forward)
0100 1101 0000 0011 1010 1100 0101 1000
0000 0000 0000 0000 0000 0000 1001 1100
.......................................................................
0100 1101 0000 0011 1010 1100 1111 0100
PC+Jump(Backwards)
0100 1101 0000 0011 1010 1100 0101 1000
1111 1111 1111 1111 1100 0000 1001 1100
.......................................................................
0100 1101 0000 0011 0110 1100 1111 0100
The offset is a signed value, so jumping backward just uses a negative offset.
The examples you've shown demonstrate simple addition. In the context of branching and jumping, the top number represents the current PC, the second number is the offset, and the bottom number is the new PC. Recall that in two's complement, subtraction is the same as addition of a negative number. Notice that in the backward example, the offset is negative. When the offset is encoded in a MIPS instruction, it gets sign-extended to the full 32 bits.