Assign 2 - Exercise 2.3

Now we are doing things slightly differently, and bringing in the handling of negatives.

Since a good portion of this is still repeated, I am removing load/store instructions form the writeup because they will not change from the previous two examples.  I am however going to continue my ARM conversions.

r0==f

r1==g

r2==h

r3==i

a. f = –g – f;

b. f = g + (–f – 5);

2.3.1 For the C statements above, what s the corresponding MIPS

assembly code? Use a minimal number of MIPS assembly instructions.

a)

MIPS	Comment	ARM
sub $r1, $zero, r1 sub $r0, $r1, $r0	#g = 0 - g #f = g - f	sub r1, #0, r1 sub r0, r1, r0

b)

MIPS	Comment	ARM
sub $r0, $zero, r0 addi $r0, $r0, -5 add $r0, $r1, $r0	#f = 0 - f #f = f-5 #f = g+f	sub r0, #0, r0 sub r0, r0, #5 add r0,r1,r0

2.3.2 For the C statements above, how many MIPS assembly instructions are needed to perform the C statement?

a) 2 subtractions = 2 Instructions

b) 1 subtract, 1 add immediate, 1 add = 3 instructions

2.3.3 I the variables f, g, h, i, and j have values 1, 2, 3, 4, and 5, respectively, what s the end value o f?

f=1 g=2 h=3 i=4

a)g = 0 - 2 = -2

  f = -2 - 1 = -3

b)f = 0-1 = -1

  f = -1-5 = -6

  f = 2 + -6 = -4

---

a. addi f, f, 4

b. add  f, g, h

   sub  f, i, f

Converted to ARM

a. add f, f, #4

b. add f,g,h

   sub f,i,f

2.2.4 [5] <2.2> For the MIPS assembly instructions above, what is a corresponding C statement?

a) f += 4

b) f = g+h

   f = i-f

f = i - (g + h)

2.2.5 [5] <2.2> I the variables f, g, h, and i have values 1, 2, 3, and 4, respectively, what s the end value of f?

f=1 g=2 h=3 i=4

a) f = 1 + 4  = 5

b) f = 2 + 3 = 5

   f = 4 - 5 = -1

I could find no $zero equivalent in ARM, and found no example of a negative being added in ARM, so I am assuming you must use sub instead.  This exercise went extremely quickly as, while adding another instruction, is still the same stuff gone over in the previous two exercises.