[SOLVED] CSE3038 HW1 - Key

1. For the following C statement, what is the corresponding MIPS assembly code? Assume that the variables f, g, h, i, and j are assigned to registers $s2, $s3, $s4, $s5, and $s6, respectively. Assume that the base address of the arrays A and B are in registers $s0 and $s1, respectively. B[i+3] = A[i+4*j];

Answer: 

2. Show how the value 0xcabd1f2e would be arranged in memory of a little-endian and a big-endian machine. Assume the data is stored starting at address 0.

Answer: 

Little Endian

Big Endian

3. Translate the following C code to MIPS. Assume that the variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3, and $s4, respectively. Assume that the base address of the arrays A and B are in registers $s6 and $s7, respectively. Assume that the elements of the arrays A and B are 4-byte words: B[i+j+1] = A[i+j-2] + A[i-j+1]; Answer:

sub $t0, $s3, $s4                     # $t0 = i – j add $t1, $s3, $s4                  # $t1 = i + j

addi $t0,$t0,1                          # $t0 = i – j+1

addi $t2,$t1,1                          # $t2 = i + j + 1

addi $t1,$t1,-2                         # $t1 = i + j-2

sll $t0, $t0, 2                            # $t0 = (i-j+1)*4

sll $t1, $t1, 2                            # $t1 = (i+j-2)*4

add $t3, $s6, $t0                       # $t3 = &A[i-j+1]

add $t4, $s6, $t1                       # $t4 = &A[i+j-2]

lw $t5, 0($t3)                           # $t5 = A[i-j+1]

lw $t6, 0($t4)                           # $t6 = A[i+j-2]

add $t7, $t5, $t6                      # $t7 = A[i+j-2] + A[i-j+1]

sll $t2,$t2,2                             # $t2 = (i+j+1)*4

add $t2,$t2,$s7                        # $t2 = &B[i+j+1]

sw $t7, 0($t2)                          # B[i+j+1] = $t7

4. Provide the type, assembly language instruction, and binary representation of instruction described by the following MIPS fields:

op=0, rs=5, rt=8, rd=20, shamt=0, funct=36.

Answer: 

Its R type instruction

Op=0 funct=36 è and

and $20, $5, $8            or

and $s4, $a1, $t0

5. For the following C statement, write a minimal sequence of MIPS assembly instructions that does the identical operation. Assume $t1 = A, $t2 = B, and $s1 is the base address of C.

A = C[0] << 8;

Answer: 

lw $t3, 0($s1) sll $t1, $t3, 8

6. Translate the following C code to MIPS assembly code. Use a minimum number of instructions. Assume that the values of a, b, i, and j are in registers $s0, $s1, $t0, and $t1, respectively. Also, assume that register $s2 holds the base address of the array D.

for(i=0; i<a; i++)  for(j=0; j<b; j++)

D[2*i] = i + j-5; Answer: 

addi $t0, $0, 0 beq $0, $0, TEST1 LOOP1: addi $t1, $0, 0 beq $0, $0, TEST2 LOOP2: add $t3, $t0, $t1

sub $t3,$t3,5 sll $t2, $t0, 3 add $t2, $t2, $s2 sw $t3, ($t2) addi $t1, $t1, 1 TEST2: slt $t2, $t1, $s1 bne $t2, $0, LOOP2 addi $t0, $t0, 1

TEST1: slt $t2, $t0, $s0 bne $t2, $0, LOOP1

7. How many MIPS instructions does it take to implement the C code from Exercise 2.27 ? If the variables a and b are initialized to 10 and 1 and all elements of D are initially 0, what is the total number of MIPS instructions that is executed to complete the loop?

Answer: 

14 instructions to implement and 144 instructions executed

Inner loop will execute ones since b is 1