While reading Computer Architecture by Patterson (page 194) I got this question. IEEE 754 uses 127 as bias for single precision floating point so that it will be easy to compare floating point numbers using hardware.

  1. 1.0 * 2^-128
    biased exponent = -128+127 = -1 = 1111 1111

  2. 1.0 * 2^1
    biased exponent = 1+127 = 128 = 1000 0000

For hardware compare 1) comes out to be a larger number than 2) which is wrong. Can anybody explain how comparing is working for these two numbers?


The exponent is unsigned. It cannot be -1. The smallest it can be is 0, which translates to 2^-127.

You might ask then, "If so, how is 2^-128 represented in single precision?!".

The answer lies in normalization.

Usually numbers are normalized (i.e. multiplied by a power of 2) so that the leading bit of their mantissa is 1. That bit can then be omitted, giving you extra precision (or removes redundancy, depending on who you ask).

It was decided for ~reasons~ (to enable a gradual loss of precision during underflow), that it's not worthwhile keeping to this schema when the exponent is -127. Rather the leading bit is whatever the leading bit is.

What this does, is it decreases resolution in the range 2^-127 to 2^-126 but allows to represent additional numbers below 2^-127.

Hence, 2^-128 will have an exponent of 0 (translated to -127 via the bias), and a mantissa of 010000...000. Note that, unlike all normalized numbers, the mantissa means 0.5 - which is outside the usual range of the mantissa for normalized numbers (1.0 to 2.0).

With this in mind - the exponent comparison makes easy sense - 128 is larger than 0 :)

All this is also given in the wiki article on IEEE 754-1985 (and probably others) in the very first section.

  • That does help.And In The book its written that pattern of all one 1111 1111 is reserved for (+-) infinity. and I think we even can't denote 1 * 2^-128 with only 32 bits in hand. Upto 1 * 2^-126 we can denote it without any problem . – prashant singh Jul 27 '15 at 3:13
  • @prashantsingh Indeed, an exponent of 1111 1111 is reserved - for infinity if the mantissa is 0, or NaN in case it's non-zero. If floating point numbers were always normalized then indeed, 2^-128 would not be possible to represent in this format. I found this site useful when answering this question. – Ordous Jul 27 '15 at 11:59

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