Floating Point Arithmetic, Part One

Floating Point Arithmetic, Part One

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A month ago I was discussing some of the issues in integer arithmetic, and I said that issues in floating point arithmetic were a good subject for another day. Over the weekend I got some questions from a reader about floating point arithmetic, so this seems like as good a time as any to address them.

Before I talk about some of the things that can go terribly wrong with floating point arithmetic, it's helpful (and character building) to understand how exactly a floating point number is represented internally.

To distinguish between decimal and binary numbers, I'm going to do all binary numbers in blue fixed-width.

Here's how floating point numbers work.  A float is 64 bits.  Of that, one bit represents the sign: 0 is positive, 1 is negative.  

Eleven bits represent the exponent.  To determine the exponent value, treat the exponent field as an eleven-bit unsigned integer, then subtract 1023.  However, note that the exponent fields 00000000000 and 11111111111 have special meaning, which we'll come to later.

The remaining 52 bits represent the mantissa.

To compute the value of a float, here's what you do.  You take the mantissa, and you stick a "1." onto its left hand side.  Then you compute that value as a 53 bit fraction with 52 fractional places.  Then you multiply that by two to the power of the given exponent value, and sign it appropriately.

So for example, the number -5.5 is represented like this: (sign, exponent, mantissa)

(1, 10000000001, 0110000000000000000000000000000000000000000000000000)

The sign is 1, so its a negative number.  The exponent is 1025 - 1023 = 2.  Put a 1. on the top of the mantissa and you get 1.0110000000000000000000000000000000000000000000000000 = 1.375 and sure enough, -1.375 x 22 = -5.5

This system is nice because it means that every number in the range of a float has a unique representation, and therefore doesn't waste bits on duplicates. 

However, you might be wondering how zero is represented, since every bit pattern has 1. plunked onto the beginning.  That's where the special values for the exponent come in.  If the exponent is 00000000000, then the float is considered a "denormal".  It gets 0. plunked onto the beginning, not 1., and the exponent is assumed to be -1022.  This has the nice property that if all bits in the float are zero, it's representing zero. Note that this lets you represent smaller numbers than you would be able to otherwise, as we'll see, though you pay the price of lower precision.  Essentially, denormals exist so that the chip can do "graceful underflow" -- represent tiny values without having to go straight to zero.

If the exponent 11111111111 and the fraction is all zeros, that's Infinity.  If the exponent is 11111111111 and the fraction is not all zeros, that's considered to be Not A Number -- this is a bit pattern reserved for errors.

So the biggest and smallest positive normalized floats are

(0, 11111111110, 1111111111111111111111111111111111111111111111111111)

which is 1.1111111111111111111111111111111111111111111111111111 x 21023, and

(0, 00000000001, 0000000000000000000000000000000000000000000000000000)

which is  1.000 x 2-1022

The biggest and smallest positive denormalized floats are

(0, 00000000000, 0000000000000000000000000000000000000000000000000001)

which is 0.0000000000000000000000000000000000000000000000000001 x 2-1022  = 2-1074, and

(0, 00000000000, 1111111111111111111111111111111111111111111111111111)

which is 0.1111111111111111111111111111111111111111111111111111 x 2-1022

Next time: floating point math is nothing like real number math.

  •   Mat Hall said:  "....n a PDP-10..".

    Do you have a PDP-10 to show proof of that????

    If you are interested in old DEC hardware [I have a PDP-8 a few PDP-11's and Vaxen], please drop me a note: david.corbin@dynconcepts.com

  • eric - forgive me, but can you explain how you 'compute that value as a 53 bit fraction with 52 fractional places' .. thanks

  • @kerry,

    Because there at 52 "actual" bits + one "implied" bit at the beginning. Go back and look at the example:

    011000.. is  1.01100...

  • what is the hardware difference between decimal floating point arithmetic(DFA) and Binary FA.

    because in actual decimal no are also stored as 1's and 0's format howthat differs and helpful over BFA

  • can anyone tell me what is the output when input is qnan or snan

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