CS2100 Computer Organisation Notes
· 8 min read
P.S. The following information is based on various resources from the course CS2100 NUS, certain parts are adapted to include my comments and explanations for my better understanding & review. I will try to include source in relevant section as much as possible to honor the authors' work (Mainly Prof Aaron & Prof Colin & TA Alvin).
P.P.S Half way through the course, I decided to change the way I take notes and therefore the information contained here only cover the first few chapters of the module.
C
- Read Array
// while-loop
int readArray(int arr[], int limit) {
int index, input;
printf("Enter up to %d integers, terminating with a negative integer.\n", limit);
index = 0; // Must remember to initialize to zero
scanf("%d", &input);
while (input >= 0) {
arr[index] = input;
index++;
scanf("%d", &input);
}
return index; // Which is the actual array size
}
// for-loop
int readArray2(int arr[], int limit) {
int index, input;
printf("Enter up to %d integers, terminating with a negative integer.\n", limit);
for (index = 0; index < limit; index ++) {
if (arr[index] < 0) {
break;
}
scanf("%d", &arr[index]);
}
return index - 1; // Which is the actual array size
}
- Reverse Array
// iterative while-loop
void reverseArray(int arr[], int size) {
int left = 0, right = size - 1, temp;
while (left < right) {
temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
left++;
right--;
}
}
// iteractive for-loop
void reverseArray2(int arr[], int size) {
int start, temp, end = size - 1;
for (start = 0; start < size / 2; start++) {
temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
end--;
}
}
// recursive
void reverseArray3(int arr[], int size) {
reverseArrayRec(arr, 0, size - 1);
}
void reverseArrayRec(int arr[], int left, int right) {
int temp;
if (left < right) {
temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
reverseArrayRec(arr, left + 1, right - 1);
}
}
// recursive C-style : from TA Alvin CS2100
void reverseArray4(int arr[], int size) {
if (size <= 1) {
return;
} else {
temp = arr[0];
arr[0] = arr[size - 1];
arr[size - 1] = temp;
// Add 1 to array pointer => shift pointer by 4 bytes
// => move to next item in array, that is the new base
// => treat it as removing the front and back items,
// hence size - 2
reverseArray4(arr + 1, size - 2)
}
}
Sign extension
When we want to represent an n-bit signed integer as an m bit signed integer, where m > n. We do this by copying the sign-bit of the n-bit signed m - n times to the left of the n-bit number to create an m-bit number.
To show that sign extension is value-preserving, we can simply check two cases:
- If the signed bit is 0, adding 0s to the left is similar to adding 0 to the right of a decimal point. It does not change the value.
- If the signed bit is 1, adding 1s to the left seems to change the value, but not really. This is because the newly added 1s to the right of the newly added signed bit increases the value while the signed bit decreases the value of the number. This effectively cancels out the changes in value. E.g. Focusing on the value that signed bit and the extended signed bit portion, original: 1001 => -2**3 = -8, extended: 111001 => -2**5 + 2**4 + 2**3 = -8
Addition of two 1's complement number with decimal
- Sign extend or pad zero to the end so that they are of equal length.
- Invert all the bits for negative number to use addition.
- If there is a carry out, add 1 to the result.
- Check for overflow if the result is opposite sign of A and B.
Converting decimal numbers to fixed-point binary
When doing the rounding
- 0.0000 is rounded to 0.000.
- 0.0001 is rounded to 0.001.
Represent decimal value in IEEE 754 format
- When converting the 32 bits to Hexadecimal, make sure that the signed bit is also included.
Bitwise operations
a = 001010
b = 101011
a | b = 101011
a & b = 001010
a ^ b = 100001
~a = 110101
a << 2 = 101000 # equivalent to a *= 4
a >> 2 = 001010 # equivalent to floor(b /= 4)
Special thing about XOR
x ^ x = 0
x ^ 0 = x
a = 00110
a = 00110
a ^ a = 00000
b = 00000
c = 00110
c ^ b = 00110
Swap without temporary variable, with bitwise operator
void swap(int *a, int *b) {
*a = *a ^ *b
*b = *a ^ *b # equivalent to *a ^ *b ^ *b = *a
*a = *a ^ *b # equivalent to *a ^ *b ^ *a = *b
}
MIPS Masking
Summary
- andi/ori/xori use zero-extension to fill in upper 32 bits. The operation acts on all 32 bits.
andi x, 1 = x
andi x, 0 = 0
ori x, 0 = x
ori x, 1 = 1
nor x, 0 = ~x
xor x, 1 = ~x
xor x, 0 = x
Set bits 2, 8, 16 to 1 in b, a -> $s0, b -> $s1, c -> $s2
# 16th bit is outside the range of immediate, which is only 0 - 15th bit
# use intermediate registers for results
lui $t0, 0b1 # this sets the 16th bit
ori $t0, 0b0000000100000100 # this sets 0 - 15th bit
or $s1, $s1, $t0
Copy over bits 1, 3 of b into a
lui $t1, 0b1111111111111111
ori $t1, t1, 0b1111111111110101
and $s0, $s0, $t1 # clear the two bits from a
andi $t0, $s1, 0b0000000000001010 # get the two bits from b
or $s0, $s0, $t0 # copy the bits over
Make bits 2, 4 of c inverse of bits 1, 3 of b
xori $t0, $s1, 0b0000000000001010 # this invert and copy
andi $t0, $t0, 0b0000000000001010 # clear everything else
sll $t0, $t0, 1 # shift left to 2, 4
lui $t1, $t1, 0b1111111111111111 # clear out 2, 4 of c, 3 steps, lui, ori, and
ori $t1, $t1, 0b1111111111101011
and $s2, $s2, $t1
or $s2, $s2, $t0 # copy over to c
MIPS arithmetic
a -> $s0 b -> $s1 c -> $s2 d -> $s3
d = 6a + 3(b - 2c) d = 6a + 3b - 6c d = 3(2a + b - 2c) d = 3(2(a - c) + b)
sub $t0, $s0, $s2 # a - c
sll $t0, $t0, 1 # 2(a - c)
add $t0, $t0, $s1 # 2(a - c) + b
sll $s3, $t0, 2 # 4(2(a - c) + b)
sub $s3, $s3, $t0 # 3(2(a - c) + b)
MIPS tracing
add $t0, $s0, $ zero # make a copy
lui $t1, 0x8000 # set t1 to 1 at MSB and 0 else where
lp: beq $t0, $zero, e # if t0 == zero, exit
andi $t2, $t0, 1 # t2 left with LSB
beq $t2, $zero, s # if t2 LSB == 0, s
xor $s0, $s0, $t1 # invert MSB
s: srl $t0, $t0, 1 # discard LSB
j lp # loop back
e: # exit
What happens when integer overflow in C
- int is 4 bytes (in sunfire), which is 4 x 8 = 32 bits
- the range is from -2,147,483,648 (-2^31) through +2,147,483,647(2^31 - 1)
- This range comes from 32 bits 2s complement representation,
- from 10000000000000000000000000000000 (smallest negative number)
- to 01111111111111111111111111111111 (largest positive number)
- when adding 1 to the largest positive number, it becomes the smallest negative number
- adding another 1 will make it 10000000000000000000000000000001,
- which is -2147483647, (literally add 1 to the decimal number)
For an n-bit sign-and-magnitude representation, what is the range of values that can be represented?
- -(2^(n-1) - 1)
- 2^(n-1) - 1
1s Complement
- -x = 2^n - 1 - x (in decimal)
- -x = 2^n - 1 - x + 1 (in decimal)
- minus 1 because you want the largest possible values in that base
2s Complement
- -2^(n - 1)
- 2^(n-1) - 1
- Adding A + B
- Ignore the carry out of the MSB
- Check for overflow. Overflow occurs if the carry in and the carry out of the MSB are different, or if result is opposite sign of A and B