1.4.1 Data Types

• Real / Floating Point – Stores decimal numbers (3.141) • Character - A single letter, number or special character (‘H’) • String – A collection of characters (“Hello World”) • Boolean – TRUE or FALSE • Integer – A positive or negative whole number (24, -34)

The process of changing one data type into another.

• A list of the characters the computer can represent. • Each character is represented by a unique binary value. • Used to map binary values to characters. Examples: UNICODE and ASCII

• ASCII is a character set which is a subset of UNICODE • Uses 7 bits, or 8 bits for extended ASCII • Fewer characters can be represented than UNICODE • Characters from different languages cannot be represented in ASCII

• Each character is represented by 1-4 bytes. • It supports a very large number of characters • It is backwards compatible with ASCII • Text using UNICODE rather than ASCII would take up more storage (roughly 4 times more)

• Use the binary number line (128 | 64 | 32 | 16 | 8 | 4 | 2 | 1) • 128+32+16+1 = 177 Answer = 10110001

Use the binary number line (128 | 64 | 32 | 16 | 8 | 4 | 2 | 1) 128+32+16+2 = 178

• Binary = 010111100 • Split from right to left into nibbles = 1011 1100 • Convert to denary using 8421 number line for each nibble • 1011 = 11 | 1100 = 12 • Remember A = 10 | F = 15 Answer = BC

• Remember A = 10 | F = 15 • F = 15 | E = 14 • Write each character in a 4-bit nibble = 1111 1110 • Combine into one binary number = 11111110 • Convert to denary = 128+64+32+16+8+4+2 Answer = 254

• Write out the positive version of 49 in binary = 00110001 • Flip and add one to the bits (2’s Complement) = 11001111 Answer = 11001111 Alternative methods exist using (-128) + 64 + 8 + 4 + 2 + 1 = -49

• No flipping required as it is a positive number • You only need to flip when it requires a negative number Answer = 00110001

• Your left-most bit (most significant) is now only available to store a 1 for negative OR 0 for positive. • Use the remaining bits to make the number (-) 0110001 Answer = 10110001

Stack the numbers on top of each other and use the rules of binary addition going from right to left • 0 + 0 = 0 • 1 + 0 = 1 • 1 + 1 = 0 (carry 1) • 1 + 1 + 1 = 1 (carry 1) 01101010 00111111 + Answer = 10101001 Carries - 11111100

• Convert the second number to a negative version (flip+add 1) • Stack the original first number on top of the new negative second number • Follow the rules of binary addition Answer = 00011000 An overflow should always occur using this method An alternative borrowing method is also suitable

• A normalised number can only start with 01 if positive, or 10 if it is negative • Remove the excess bits from the front of the number - be careful not to change the sign of the number. (01011000) • For each digit removed - pad the right of the mantissa with 0s (0101100000) • Convert the original exponent to denary and subtract the number of number bits that you removed from the front of the mantissa (12-2) • Convert the exponent back to binary (it could be a negative number) Answer = 0101100000 001010 (Remove the two extra bits from the front of the mantissa and reduce the exponent value by 2) Alternative methods are available

• A normalised number can only start with 01 if positive, or 10 if it is negative • Remove the excess bits from the front of the number - be careful not to change the sign of the number (10011000) • For each digit removed - pad the right of the mantissa with 0s (1001100000) • Convert the original exponent to denary and subtract the number of number bits that you removed from the front of the mantissa (12-2) • Convert the exponent back to binary (it could be a negative number) Answer = 1001100000 001010 (Remove the two extra bits from the front of the mantissa and reduce the exponent value by 2) Alternative methods are available

• Write out the number in fixed point binary (0101.01) • Float the binary point to be to the left of the first 1 (0.10101) • Write out the number without the binary point (010101) - this is your mantissa. If it is not long enough then pad on the right with zeros. • In binary, write out how many places you floated the binary point (011) this is your exponent. If it is not long enough then pad on the left with zeros. Answer = 010101 011 Alternative methods are available

• Write out the number in fixed point binary (0101.01) • Float the binary point to be to the left of the first 1 (0.10101) • Write out the number without the binary point (010101) - this is your mantissa. If it is not long enough then pad on the right with zeros. • In binary, write out how many places you floated the binary point (011) this is your exponent. If it is not long enough then pad on the left with zeros. • We need the mantissa to be negative - so flip and add one to the bits. Answer = 101011 011 Alternative methods are available

Pause! look at the binary. It starts with a zero, so is postive.. the exponent also starts with a zero, so overall this is not a really tiny number. * Convert the exponent (0001) to denary (Exp = 1) * If the mantissa is negative then flip and add one to it. * Add the binary point to the left of the first 1 in the mantissa (0.1110) * Float the binary point as dictated by the exponent - left if negative exponent OR right if positive exponent (01.110 in fixed point). * Calculate the denary value - normal binary to the left of the binary point and fractions to the right of it (1/2 | 1/4 | 1/8 | 1/16) Answer = 1.75 Alternative methods are available

Pause! look at the binary. It starts with a one, so is negative..The exponent starts with a zero, so overall this is not a really tiny number. • Convert the exponent (0001) to denary (Exp = 1) • If the mantissa is negative then flip and add one to it (01110) • Add the binary point to the left of the first 1 in the mantissa (0.1110) • Float the binary point as dictated by the exponent - left if negative exponent OR right if positive exponent (01.110 in fixed point). • Calculate the denary value - normal binary to the left of the binary point and fractions to the right of it (1/2 | 1/4 | 1/8 | 1/16) Answer = -1.75 Alternative methods are available

• Write out the number in fixed point (0.0010) - 1/8 • Calculate how many places the binary point needs to float to be to the left of the left-most one (2 places to the right, this is a negative movement) • Remove the additional zeros from the front of the mantissa to get it to be 01 for positive or 10 for negative. • Pad the mantissa on the right hand side to make it to the required number of bits (010000) • Write out your exponent 010 - flip to be negative 110 Answer = 010000 110 Any number less than 0.5 would have a negative exponent

• Ignore the negative - write out the number in fixed point (0.0010) - 1/8 • Calculate how many places the binary point needs to float to be to the left of the left-most one (2 places to the right, this is a negative movement) • Remove the additional zeros from the front of the mantissa to get it to be 01 for positive or 10 for negative. • Pad the mantissa on the right hand side to make it to the required number of bits (010000) • Flip and add one to the mantissa to make it negative. • Write out your exponent 010 - flip to be negative 110 Answer = 110000 110 Alternative methods are available

• Allows for more accuracy/precision from the given number of bits • So that the representation of each binary value is unique

Increases the size of the number that can be stored. Allows us to move the binary point further