Understanding exponent 00000000 and 11111111 in IEEE

Question

I can't get my head around these exponents:

• Why is exponent 0000 0000 = -126 instead of -127 (i.e. 0 - 127)?
  • What is the exact value of exponent 1111 1111?
  • Why does 0 1111 1111 0000 0000 0000 0000 0000 000 equal infinity?

I'm very new to all this, so please explain in simple lay terms! Thanks!

Answer 1

Why is exponent 0000 0000 = -126 instead of -127 (i.e. 0 - 127)?

For most normal exponents, when we reach the bottom of the significand range, we step down to the next exponent. For example, the representable numbers from 1 (inclusive) to 2 (exclusive) are, in descending order:

• 1.11…111 • 2⁰,
• 1.11…110 • 2⁰,
• 1.11…101 • 2⁰,
• 1.11…100 • 2⁰,
• …
• 1.00…011 • 2⁰,
• 1.00…010 • 2⁰,
• 1.00…001 • 2⁰,
• 1.00…000 • 2⁰.

Then, when going to the next lower numbers, we adjust the exponent, so the next lower numbers are:

• 1.11…111 • 2⁻¹,
• 1.11…110 • 2⁻¹,
• 1.11…101 • 2⁻¹,
• 1.11…100 • 2⁻¹,
• …

Now, when we are at the lowest normal exponent, the lowest numbers in this range are:

• 1.00…011 • 2⁻¹²⁶,
• 1.00…010 • 2⁻¹²⁶,
• 1.00…001 • 2⁻¹²⁶,
• 1.00…000 • 2⁻¹²⁶.

In order to keep going at this point, it was decided the final step would be to change the leading bit to 0 instead of 1. There was a choice here: Numbers with a zero in the exponent field could continue the pattern, going on to exponent −127, so the next representable number would be 1.11…111 • 2⁻¹²⁷, or they could stick at exponent −126 but change the leading bit to 0 instead of 1.

If the pattern had continued, then, for example, 1.11…111 • 2⁻¹²⁷ and 1.11…110 • 2⁻¹²⁷ would both be representable numbers, but their difference, 0.00…001 • 2⁻¹²⁷ = 1 • 2⁻¹⁵⁰, would not be representable. So, if x and y were these numbers, their computed difference, x-y, would have to be 0 due to rounding (when rounding to the nearest representable value). However, sometimes people wrote code like this:

if (x == y)
    Handle special case.
else
    Handle normal case with some calculation involving division by x-y.

So continuing the exponent pattern to −127 would break some code and cause floating-point arithmetic to act in undesired ways. So, the choice was made to break the pattern, keep the exponent at −126, and make the leading bit 0 instead of 1. Then the next representable numbers are:

• 0.11…111 • 2⁻¹²⁶,
• 0.11…110 • 2⁻¹²⁶,
• 0.11…101 • 2⁻¹²⁶,
• 0.11…100 • 2⁻¹²⁶,
• …
• 0.00…011 • 2⁻¹²⁶,
• 0.00…010 • 2⁻¹²⁶,
• 0.00…001 • 2⁻¹²⁶,
• 0.00…000 • 2⁻¹²⁶.

You can see that we want 0.11…111 • 2⁻¹²⁶ just after 1.00…000 • 2⁻¹²⁶. If it were 0.11…111 • 2⁻¹²⁷ instead, there would be a gap—that is half the size it needs to be.

What is the exact value of exponent 1111 1111?

When the exponent field is 1111 1111, it does not represent any numeric exponent for the normal floating-point format. That value in the exponent field is a code for special values (infinity and NaN).

Why does 0 1111 1111 0000 0000 0000 0000 0000 000 equal infinity?

There is not a mathematical reason for this. Infinity does not arise out of any mathematics performed on the significand and the exponent. It was just decided that the exponent field 1111 1111 would represent infinities and NaNs, and that, when the significand field is all zeros, it represents infinities, and, when the significand field is not all zeros, it represents a NaN.

Answer 2

The exponent bias for a single-precision float is 127. This means that an exponent value of 127 (1000 0000) is evaluated as an actual exponent of zero, and an exponent value of one (0000 0001) is evaluated as an actual exponent of -126.

Exponent values of all zero bits or all one bits (0000 0000 or 1111 1111) are not actual exponents, but instead trigger special case behavior. An all-zero exponent means there is no implied leading one bit in the mantissa; this is used for subnormal numbers. And an all-one exponent is used to mark that the value is either infinity, negative infinity, or NaN, depending on the sign bit and mantissa bits.

For further reading, The Floating Point Guide has a nice explanation in easy terms, and of course if you really want the nitty-gritty, nothing beats Goldberg's paper.