Floating point, the OCR way

Mantissa and exponent, both in two's complement. Toggle the bits, type a number, then step through each conversion the way you would write it in the exam. A Level Computer Science, OCR H446, section 1.4.1 (e) and (f).

Mantissa bits Exponent bits Quiz me

value = mantissa × 2exponent

Example with 8 and 4 bits: 0.1011000 × 2⁰⁰¹¹ = 0.6875 × 8 = 5.5

The mantissa holds the significant bits. Its binary point sits immediately after its first bit, and that first bit is the sign.
The exponent is an ordinary two's complement integer: how many places to move the point (right if positive, left if negative).
Normalised means the mantissa starts 0.1 (positive) or 1.0 (negative). The first two bits differ.
The question tells you the bit widths. Always answer with that many bits.

Floating point number

Mantissa bits

Exponent bits

Exam method

Write the mantissa with its point after the first bit; write the exponent separately.
First bit of the mantissa is the sign. If it is 1, find the two's complement to get the size of the number.
Convert the exponent (two's complement) to denary.
Move the point that many places: right for a positive exponent, left for a negative one.
Add up the place values of the 1s (… 4, 2, 1, ½, ¼, ⅛ …).
Put the sign back.

Worked conversion

Denary number

Number to convert

Exam method

Note the sign. Work with the positive value.
Integer part to binary (place values, or repeated division by 2).
Fraction part to binary by repeated doubling: the digit before the point each time is the next bit.
Normalise: move the point to just before the first 1. Each place moved left is +1 on the exponent; each place moved right is −1.
Mantissa: a 0, then the bits after the point, padded with zeros (or truncated) to the right width.
Exponent: two's complement in the given width.
Negative number: two's complement the whole mantissa. If it now starts 1.1, shift it left once and subtract 1 from the exponent.

Worked conversion

Unnormalised number

Mantissa bits

Exponent bits

Exam method

A normalised mantissa starts 0.1 or 1.0: the first two bits differ.
If the first two bits are the same, shift the mantissa one place left (a 0 comes in on the right) and subtract 1 from the exponent.
Repeat until the first two bits differ.
Check: convert before and after to denary; the value must not change.

Why normalise?

One representation per number. 0.011 × 2³ and 0.11 × 2² are both 3; normalised form makes the second one the only correct answer, so numbers can be compared bit by bit.
Maximum precision. Leading bits that merely repeat the sign are wasted. Shifting them out frees those bit positions for real significant bits.

Worked normalisation

Two numbers

A (denary)

Operation

B (denary)

Exam method

Make the exponents equal: shift the mantissa with the smaller exponent to the right by the difference, bringing copies of its sign bit in on the left, and raise its exponent to match.
To subtract, two's complement the mantissa being subtracted, then add.
Add the mantissas as two's complement numbers. Leave a spare bit on the left for the carry, because the result can be up to twice the size.
Normalise the result: shift right (exponent +1) if it has grown past the point, or shift left (exponent −1 each time) until it starts 0.1 or 1.0.
Truncate the mantissa to the required width and check the exponent still fits.

Worked calculation

Exam tips and common mistakes

Both parts are two's complement. The exponent is not sign and magnitude and there is no bias. If an exponent starts with 1, it is negative.
The point is after the sign bit, not before it. 1.011 as a negative mantissa is −1 + ¼ + ⅛ = −0.625, not −0.375.
Pad to the full width. A 4-bit exponent of 3 is 0011, not 11. An 8-bit mantissa for 0.1 is 01000000.
Negative numbers: convert the positive value first, take the two's complement of the whole mantissa, then re-check normalisation. −0.5, −1, −2 and −4 all end up with mantissa 1.000… and need the exponent reduced by one.
Moving the point. A positive exponent moves it right (number gets bigger). Students often go the wrong way; check that a positive exponent with a 0.1… mantissa gives a number of at least ½.
Zero cannot be normalised. By convention, mantissa 0 and exponent 0.
Show the working. Marks are given for the two's complement, the doubling table, the shift count and the exponent, not just the final bits. Either method for negative mantissas (two's complement first, or a place value of −1 for the sign bit) is accepted.
Not on the spec: real processors use the IEEE 754 standard, with a biased exponent and a hidden leading 1. OCR questions never use it, so do not mix the two conventions.

Mantissa: The part of a floating point number that holds the significant digits. In OCR's form it is a two's complement fraction with the point after the sign bit.
Exponent: The power of two the mantissa is multiplied by; in effect, how far the point moves.
Normalised: A mantissa that starts 0.1 or 1.0, giving each value a single representation and the maximum number of significant bits.
Truncation: Dropping the bits that do not fit in the mantissa. The stored value is then slightly closer to zero than the true one.

Floating point, the OCR way

Floating point number

Exam method

Worked conversion

Denary number

Exam method

Worked conversion

Unnormalised number

Exam method

Why normalise?

Worked normalisation

Two numbers

Exam method

Worked calculation

Range and precision

Trading bits: every split of the same total

What to say in the exam

Practice