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Converting Mips to binary 
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#1
Jan3111, 06:42 PM

P: 18

Hey Guys!!!! Thank you for checking out my Thread.... I am taking a computer assembly class and i have homework due on Wed but i am stuck on a single problem. I have to convert MIPS instructions into binary, I want to learn how to do this because it will be on a test... and i could just go into mars and assemble the code and read it from there but i wanna know how to do it..... okay soo the instructions are:
addi $sp, $sp, 8 sw $ra, 4($sp) Now from looking into my book i see that addi has an Opcode of 001000 and it has a I instruction format... Also sw has an Opcode of 101011 and is an I instruction format How do i figure out what registers $sp?, $ra, the array 4($sp) are in binary? Please help... i would really appreciate it! 


#2
Jan3111, 06:59 PM

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P: 21,397

4($sp) would be an address on the stack  $sp + 4*(size of a machine word). Here's another link to a site that lists the instructions and what they look like in binary  http://www.mrc.uidaho.edu/mrc/people...al/MIPSir.html. For your ADDI instruction, the constant to be added is 8, which will need to be written in two's complement form. 


#3
Jan3111, 07:19 PM

P: 18

Thank you for your reply... i have already checked those sites out... but they have not helped me at all....
So how do i go about converting it into binary code? i think i am missing the simple concept... So i take Addi for instance and use its opcode... okay cool got that done but what about the registers? all it says on that web site is: Used to add signextended constants (and also to copy one register to another: addi $1, $2, 0), executes a trap on overflow 001000ss sssttttt CCCCCCCC CCCCCCCC what the hell does that mean lol 


#4
Jan3111, 07:47 PM

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P: 21,397

Converting Mips to binary
This 001000ss sssttttt CCCCCCCC CCCCCCCC  shows how the instruction op code and registers and operands are encoded to make up an ADDI instruction.
First 6 bits  the instruction opcode (001000 for ADDI) Next 5 bits  the source ($2 in your example) Next 5 bits  the destination ($1 in your example) Next 16 bits  the immediate value ADDI effectively does this $t = $s + imm. value So if you wanted to add 5 plus the value in $t1 to $t2, the MIPS assembly instruction would be: ADDI $t2, $t1, 5 Note that the numbers associated with $t0 and $t1 are 8 and 9. The binary form of this instuction would be



#5
Jan3111, 08:22 PM

P: 18

Thank you again for clarifying that...but what if there is no value for the registers i see that i have a constant of 8 which i would use two's complement but for the other $sp register is it defaulted at 0?



#6
Jan3111, 08:27 PM

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P: 21,397

Are you asking about ADDI $sp, $sp, 8?
What do you mean "but for the other $sp register is it defaulted at 0" 


#7
Jan3111, 08:41 PM

P: 18

Yes i am
addi $sp, $sp, 8 would that in the end turn to be (opcode) 001000 (default 0 for $sp) 00000 (default 0 for $sp) 00000 1111111111110111 (8) 


#8
Jan3111, 08:52 PM

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#9
Jan3111, 09:12 PM

P: 18

oohhh i think i may understand.... so the value of $sp ( the stack pointer) is $29 sooo
it would be 001000 11101 11101 111111111111011 opcode $sp $sp 8 constant did i correctly get the value? 


#10
Jan3111, 09:14 PM

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P: 21,397

Looks good!



#11
Jan3111, 09:35 PM

P: 18

okay SWEET!!! soo and for the other one i got
101011 11111 11111 how do i figure out the 4($sp) ? you stated before "4($sp) would be an address on the stack  $sp + 4*(size of a machine word)." so do i take 29 + 4 * ( idk what the size of word is) ? 


#12
Feb111, 12:48 AM

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P: 21,397

sw $ra, 4($sp)
What this is doing is to store the word (== 32 bits) at $sp + 4 to the $ra register. The size of a machine word seems to be 32 bits, since that's the size of the encoded instruction. Let's go with that assumption. According to the reference I cited, the encoded instruction is 101011 sssss ttttt iiii iiii iiii iiii. Here, the s bits hold the source register, the t bits the destination register, and the i bits the offset from the source register. I am thinking now that the offset is in bytes, so we use the offset asis. So the number for $sp is $29, the number for $ra is $31, and the offset is 4. I think this is the encoded instruction for SW $RA, 4($SP) 101011 11101 11111 0000 0000 0000 0100 The first 6 bits are the SW instruction. The next 5 bits are the source register, $SP. The next 5 bits are the $RA register, and the last 16 bits are the offset, which is 4. 


#13
Feb111, 05:59 PM

P: 18

Thank you so much for your help!! You have really helped me understand this!! I am very grateful!!!!



#14
Feb111, 06:02 PM

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P: 21,397

You're very welcome. Thanks are always appreciated!



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