A Level Computer Science Paper 1: 1.4.1 - Data Types - Using binary

One provided by a programming language

Integer, float, char, string, Boolean

Base 10/decimal

Our number system

It will have 2/10/16 written in subscript

Uses 1/0 multiplied by the base, starting from the right

The bases are all powers of 2

Put a 1 in the column that is closest to the denary value, subtract and repeat

Uses base 16

Letters A-F represent values 10-15

Quicker
Easier to understand
Less likely to make errors

Show workings

Convert letters to digits, multiply by base and sum

Divide by 16, add the hex symbol of the remainder assuming 2 bit value

Each hex value is 4 bits of binary
Convert the hex or binary to digit and then write out as the other
Combine the outputs

Works similar to denary, carry over a 1 if you have to go back to 0

There is an overflow error and the computer just cuts off the last bit from its output
Clearly indicate this in your answers

0 -> (2^n) -1 where n is the number of bits

You make the second number into its negative value using Two’s Complement and then add them

The first bit represents +/- and then the rest are the same

The first bit being a 1 means the number is negative.

The leftmost column has the negative value of itself

-(2^(n-1) + 1 -> (2^)n-1)) -1

1. Find the positive value

2. Invert the value of each bit

3. Add one

(2^(n-1)) -> 2^(n-1) - 1

You can ignore those bits

lsb - least significant bit - the rightmost bit

msb - most significant bit - the leftmost bit

Pushes every value 1 space to the left, the msb goes into the carry bit, 0 comes into the lsb

Pushes every value 1 space to the right, the lsb goes into the carry bit, 0 comes into the msb

x2 for left, /2 for right as long as it isn’t two’s complement or sign and magnitude

Pushes every value 1 space to the left but the msb doesn't move so the second bit from the left goes into the carry bit, 0 comes in as lsb

Pushes every value 1 space to the right and the value that comes into the msb is whatever was there before

x2 for left, /2 for right, for all numbers

Pushes every value 1 space to the left, the msb into the carry bit and the carry bit into the lsb

Pushes every value 1 space to the right, the lsb into the carry bit and the carry bit into the msb

Apply another binary value with an AND, OR or XOR

| Use boolean algebra on binary values by using each pair of corresponding values

0: The output is the same as the input
1: The output is 1 (set the bit)

0: The output is 0 (clear the bit)
1: The output is the same as the input

0: The output is the same as the input
1: The output is the opposite of the input (toggle the bit)

At a fixed point is a metaphorical decimal point, after that comes negative values of 2

0.5, 0.25, 0.125, 0.0625, 0.03125

Having more bits before the decimal point increases the range but decreases the accuracy

A mantissa and an exponent

| mantissa x 2^exponent

Between the first 2 bits

1. Find the value of the exponent

2. Move the decimal point by that many spaces to the right

3. Find the value of that as if it was a fixed point decimal

1. Find the two’s complement of the mantissa

2. Find the value of the exponent

3. Move the decimal point that many spaces to the right

4. Find the value as if it was a fixed point decimal

5. Put a sign in front of your value

It will have a 1 as the msb (in two's complement)

1. Find the value of the exponent remembering the negative left value

2. Move the decimal point that many places to the left, as if there were all 0s to the left of your number

3. Find the value as if it was a regular fixed point number

Add the 1 to the lsb

Has a sign bit 0 and a 1 for the next bit

1. Move the decimal point to the right until it is in front of the 1

2. Add as many zeroes to the end as you moved right by

3. Subtract the amount of spaces you moved to the right from the exponent

4. Write out the normalised number with the mantissa and exponent

Has a sign bit 1 and a 0 for the next bit. Leading 1s don't change the value.

1. Move the decimal point to the right until it is in front of a 0

2. Add as many zeroes to the end as you moved right by

3. Subtract the amount of spaces you moved to the right from the exponent

4. Write out the normalised number with the mantissa and exponent

1. Convert it to fixed point binary

2. Move the decimal point to the left until it is in front of the first 1

3. Pad the RHS with zeroes to keep the same length if necessary

4. Set the exponent equal to how many spaces you moved to the left

1. Convert the positive number to fixed point (write a 0 at the start)

2. Take the two’s complement

3. Convert as you would a positive number

1. Convert each floating point number to fixed point binary by shifting the binary point right by the value of the exponent

2. Sum the two values

3. Convert back to floating point and normalise

1. Convert each floating point number to fixed point binary by shifting the binary point right by the value of the exponent

2. Take the two’s complement of the negative value

3. Sum the two values

4. Convert back to floating point and normalise