the question is above. and the solution to this question is below. can you pleas

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the question is above. and the solution to this question is below. can you please explain step by step on how they got the solution

3. The following codes are a set of MIPS instructions (hexadecimal form) and their corresponding addresses in the memory. For each line, the information is [Address] Instruction [00400024] 00008020 [00400028] 20110003 [0040002 cl 20120014 [00400030] 16120002 [00400034] 2231ffff [00400038] 08100010 [0040003c] 22310001 [00400040] 00118020 (1) Write down the corresponding MIPS ASM instructions (2 points x 80 (2) Write down the corresponding C codes (4 points) Solution

Explanation / Answer

(1)The corresponding MIPS ASM instructions given are..

The given instructions are given in the hexadecimal are converted in binary i.e;

binary to hexadecimal

0000 0

0001 1

0010 2

0011 3

0100 4

0101 5

0110 6

0111 7

1000 8

1001 9

1010 a

1011 b

1100 c

1101 d

1110 e

1111 f

so the instructions given are

00008020 can be written in binary as 0000 0000 0000 0000 0100 0000 0010 0000

0 0 0 0 8 0 2 0

similarly remaining instructions can be arranged i.e;

20110003 can be written in binary as 0010 0000 0001 0001 0000 0000 0000 0011

2 0 1 1 0 0 0 3

similarly you can understand remaining instructions given in the last block by comapringwith binary to hexadecimal table.

Now let us discuss the C program part as well as assembly level part

as in the C program block given

int a=0; int b=3;

if(a!=20) //since a = 0 initially. given a!=20 means a not equal 20 therfore that is true it goes to the statement b=b+1; that mens b becomes 3+1; b=4;

{

b=b+1; //=> b=3+1=4.

else

b=b-1;

}

a=b; //a=b means a becomes 4.. again a=4 goes to the loop [ if(a!=20) ] that means now a =4 which is not equal to 20. therefore now b=b+1; takes place that means b=4+1; b=5;

now after 16 times loop a=20 that means loop [ if(a!=20) ] false. therefore now b=b-1; statement takes place. now b=19 then a=b => i.e; a=19 again loop repeats.

Now we'll discuss the asm level language.

given that.

add $s0,$0,$0 # statement means s0 = 0+0 =0 initially s0=0

addi $s1,$0,03   # statement means s1 = 0+3 =3 initially s1=3

addi $s2,$0,20    # statement means s2 = 0+20 =20 initially s2=20

bne $s0,$s2, LabelA   # statement means next instruction is at LabelA if s0 not equal to s2 (i.e s0!=s2)

addi $s1,$s1,-1   # statement means [s1= s1-1 ]

j labelB  # statement means jump to LabelB

LabelA: addi $s1,$s1,$1  # statement means s1 = s1+1 =3+1 = 4; initially s1=3

LabelB: add $s0,$0,$s1  # statement means s0 = 0+1 = 1 initially s0=1

here add= addition, addi = signed addition, bne= b not equal to and j = jump.

These programs are very easy if u understand the C program it is similar to assembly level language. Both are creating a loop. In C program it is given variables as a, b whereas in Assembly language it is given as s0,s1.

I hope you have got this problem.

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