Given the following segment of a MIPS assembly code and its corresponding machin

Question

Given the following segment of a MIPS assembly code and its corresponding machine code. Fill in the blanks assuming the message1 occupies 33 characters and stored in locations 0x11110000-0x1111020 and message2 follow message1.

Location Machine Code Assembly Program

0x00400000 0x3402 ________ loop: li $v0, -7

0x00400004 0x3c04_________ la $a0, message1

0x00400008 0x1844_________ blez $v0, next

0x0040000c 0x3408_________ li $t0, 2

0x00400010 0x01024020 : add $t0,$t0,$v0

0x00400014 0x1440__________ bnez $v0, loop

0x00400018 0x3c01___________ next: la $a0, messge2

0x0040001c 0x3424_________

Can someone please explain how each assembly code translates to the corresponding machine code?

Explanation / Answer

            It is an I-type instruction ,So format is <opcode(6-bits)   rs(5 bits) rt(5 bits) immediate(16 bits)>

            addiu opcode=0x09   rs=$0 rt =$v0 immediate=2's complement of 7

           001001     00000    00000 1111 1111 1111 1001

         Convert into hex divide each bits into 4 -bits to convert

          0x2402fff9

               Convert into pseudocode--> lui $1,higher bits address

                                                             ori $a0,$1,lower bit of address

             Higher address= 00010001000100001

             Lower address= 0001000000100001

oring both we get , 0001000100110001

oring both opcodes, lui opcode=0x0f   and ori opcode=0x0d we get,001111

This is an I-type instruction,so format=<opcode(6-bits)   rs(5 bits) rt(5 bits) immediate(16 bits)>

              001111 00000 0010 0001000100110001

                0x3c041131

------------------------------------------------------------------

          blez opcode=0x06 rs=$v0 rt=$v0,offset=3("next" label cominf after 3rd instruction)

         000101   00010 00010 0000 0000 0000 0011

         0x18040003

---------------------------------------------------

       Convert into pseudocode-->addiu $t0,$0,2

       addiu opcode=0x09   rs=$0 rt =$t0 immediate=2

     001001 00000 01000 0000 0000 0000 0010

   0x24080002

        R-type instruction format <opcode(6bits-000000) rs(5 bits) rt(5 bit)rd(5bits)shampt(5 bits)func(5bits>

000000 01000 00010 01000 00000 100000

0x01024020

-------------------------------------------

               Pseudocode-->bne $v0,$v0,loop

             bne opcode=0x05

         000101 00010 00010 1111 1111 1111 1001(2's complement of 7 , because loop wiillbe backward 7th instruction ,so -7th take 2's complement

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