f = g h; b.) f = g + (h 5)

a.) f = g – h;

b.) f = g + (h – 5);

2.1.1 For the C statements above, what is the corresponding MIPS assembly code? Use a minimal number of MIPS assembly instructions.

a.) sub f, g, h             # The sum of g and h is placed in f

b.) subi f, h, 5             # Temporary variable f contains h – 5

addi f, f, g # the sum of f “(h-5)” and g in the place of f
2.1.2 For the C statements above, how many MIPS assembly instructions are needed to perform the C statement?

For f = g – h; you would only need one subtraction, so only one MIPS assembly instruction is needed. But for f = g + (h – 5); there is one addition and one subtraction, so we’ll need two MIPS assembly instructions.

The end values of f are:

f = g – h; —-> f = 2 – 3 ——> f = -1

f = g + (h – 5); ——-> f = 2 + ( 3 – 5) ——> f = 2 + (-2) —–> f = 0

even though there is a value for F to begin with, that doesn’t matter seeing as f is going to be over written.
2.1.4 For the MIPS assembly instructions below, what is a corresponding C statement?

a.) addi f, f, 4

b.) add f, g, h
add f, i, f
a.) f = f + 4;

b.) f = g + h;           // (g + h) is being stored in the value of f

f = i + (g + h);       // same as the two before but is more readable

a.) f = f + 4; —-> f = 1 + 4 —> f = 5

f = i + (g + h); —-> f = 4 + (2 + 3) —-> f = 4 + 5 —–> f = 9

a.) f = g – f;

b.) f = i + (h – 2);

2.2.1 For the C statements above, what is the corresponding MIPS assembly code? Use a minimal number of MIPS assembly instructions.

a.) sub f, g, f            # The sum of g and f is placed in f

b.) subi f, h, 2                  # Temporary varible f contains h – 2

For f = g – f; there is only one subtraction so that’s going to be only one MIPS assembly instruction. For f = i + (h – 2); there is one addition and one subtraction, so there is going to be two MIPS assembly instructions.

a.) f = g – f; —> f = 2 – 1 —> f = 1

b.) f = i + (h – 2); —> f = 4 + (3 – 2) —> f = 4 + 1 —> f = 5
2.2.4 For the MIPS assembly instructions below, what is a corresponding C statement?

a.) addi f, f, 4

b.) add f, g, h
sub f, i, f

a.) f = f + 4;

b.)  f = g + h;           // (g + h) is being stored in the value of f
f = i – f; // the value of f from the first statement is being subtracted to i and stored in f

f =  i – (g + h);       // same as the two before but is more readable

a.) f = f + 4; —-> f = 1 + 4 —> f = 5

b.) f = g + h; —> f = 2 + 3 —> f = 5
f= i – 5; —-> f = 4 – 5 —> f = -1

f = i – (g + h); —-> f = 4 – (2 + 3) —> f = 4 – 5 —> f = -1

a.) f = -g – f;

b.) f = g + (-f – 5);

2.3.1 For the C statements above, what is the corresponding MIPS assembly code? Use a minimal number of MIPS assembly instructions.

a.) sub f, -g, f              # The sum of -g and f is placed in f

b.) subi f, -f, 5           # Temporary variable f contains -f – 5

For f = -g – f it’s just one subtraction so that will be only one MIPS assembly instruction. For f = g + (-f – 5) that is one addition and one subtraction, so you will need two MIPS assembly instructions.

a.) f = -g -f; —-> f = -2 – 1 —-> f = -3

b.) f = g + (-f – 5); —> f =  2 + (-1 – 5) —-> f = 2 + (-6) —> f = -4
2.3.4 For the MIPS statements below, what is a corresponding C statement?

a.) addi f, f, -4

b.) add i, g, h
add f, i, f

a.) f = f + (-4);

b.) i = g + h; // (g + h) is being stored in the value of i
f = i + f; // the value of f from the first statement is being added to i and stored in f

f = (g + h) + f;   // same as the two before but is more readable

a.) f = 1 + (-4); —–> f = -3

b.) i = g + h; —> i = 2 + 3 —> i = 5
f = i + f; —-> f = 5 + 1 —> f = 6

f = (g + h) + f; —-> f = (2 + 3) + 1 —> f = 5 + 1 —-> f = 6

Assumptions: variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3, and $s4, respectively.

a.) f = -g – A[4];

b.) B[8] = A[i - j];

2.4.1 For the C statements above, what is the corresponding MIPS assembly code?

a.) lw   $t0, 32($s6)              # Temporary reg $t0 gets A[4]
sub   $s1, $zero, $s1            # Subtracting 0 – g to get -g, contained in $s1
sub $s0, $s1, $t0                 # $s0(f) = $s1(-g) – $t0(A[4])

b.) sub $t0, $s3, $s4              # Temporary variable $t0 gets i –j, i is stored in $s3 and J is stored in $s4

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

For both a) and b) there are 3 MIPS assembly instructions where needed to perform the C statement. Because we have to perform a (lw) transfer to a register first and then we perform a (sw) transfer back to the memory.
2.4.3 For the C statements above, how many different registers are needed to carry out the C statement?

For a) we used; $t0, $s6, $s1, $zero, $s0. Five different registers were need to carry out the C statement. For b) we used; $t0, $s3, $s4, $s1, $s6, $t1, $s7. Six different registers were needed to carry out the C statement.

Assumptions: variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3, and $s4, respectively.

a.) f = f + A[2];

b.) B[8] = A[i] + A[j];

2.6.1 For the C statements above, what is the corresponding MIPS assembly code?

a.) lw   $s0, 4($s6)              # Temporary reg $s0 gets A[2]

sub $s0, $s0, $s1 #$s0(f) = $s0(A[2]) - $s1(f)

b.) add $t0, $s6, $s3              # $t0 = $s6(A[2]) + $s3(i)

lw $s0, 4($t0)                        # B[8] = A[i] +A[j]

For a) there are 3 MIPS assembly instructions where needed to perform the C statement. For b) there are 4 MIPS assembly instructions where needed to perform the C statement. Because we have to perform a (lw) transfer to a register first and sub and add operations.

For a) we used; $s0, $s6, $s1, $s2. Four different registers were need to carry out the C statement. For b) we used; $t0, $t1, $s0, $s6, $s7. Five different registers were needed to carry out the C statement.

a.) 00000 0010 0001 0000 1000 0000 0010 0000_two

2.10.2 What type (I-type, R-type, J-type) instruction do the binary entries above represent?

a.) 0x01084020

b.) 0x02538822
2.11.1 What binary number does the above hexadecimal number represent?

Using a hexadecimal converter I got

a.) 1010 1110 0000 1011 1111 1111 1111 1100_two

b.) 1000 1101 0000 1000 1111 1111 1100 0000_two

2.11.2 What decimal number does the above hexadecimal number represent?

Using a hexadecimal converter I got

a.) 2920022012

b.) 2366177216

2.11.3 What instruction does the above hexadecimal number represent?

Using a hexadecimal converter I got

a.) sw $t3, -4($s0)