IEEE 754

numeric

Definition

IEEE 754

IEEE 754 is the dominant standard for floating-point number representation on computers, fixing $b=2$ and a family of precisions via the parameter set

$$\mathbb{F}(2,p,e_{min},e_{max},\text{denorm}).$$

Each format allocates a bit word into a one-bit sign, an exponent field of $w$ bits, and a mantissa field of $p-1$ bits (the leading $1$ of a normalised mantissa is implicit, so only the fractional $p-1$ bits are stored). The exponent is encoded with an excess (bias) $\beta =-e_{min}+1$, so the stored bits represent $e+\beta$.

Formats

The two most common formats fix the parameters as follows:

Format	$p$	$w$	$e_{min}$	$e_{max}$	$\beta$	total
binary32 (single)	24	8	$-126$	$+127$	127	32
binary64 (double)	53	11	$-1022$	$+1023$	1023	64

Conventions

Some bit-patterns are reserved and do not encode ordinary numbers.

Zero

Exponent bits all $0$, mantissa all $0$. Either sign bit yields $\pm 0$.

NaN

Exponent bits all $1$, mantissa with at least one bit set to $1$.

$+\infty$ / $-\infty$

Exponent bits all $1$, mantissa all $0$, distinguished by the sign bit.

Two further conventions handle edge cases of the normalised encoding.

Denormals

When the stored exponent is all $0$ and the mantissa is nonzero, the leading bit is taken as $0$ (not $1$), giving subnormal numbers that fill the gap between $0$ and the smallest normal number.

Gradual underflow

Denormals make the gap to zero shrink by a factor $2$ each step instead of collapsing from the smallest normal straight to zero, softening the loss of precision.

Examples

Converting $(-172.625{)}_{10}$ to IEEE 754 single precision

$\mathbb{F}(2,24,-126,+127,\text{denorm})$

Step 1. Convert to binary $(-172.625{)}_{10}$ into the binary number system:

Convert

$$\begin{array}{rl}(-172{)}_{10}& =(-10101100{)}_2\\ (0.625{)}_{10}& =(0.101{)}_2\\ & \\ ⟹(-172.625)& =(-10101100.101{)}_2\end{array}$$

Step 2. Normalise

$$(-1\underset{7\text{bits}}{\underbrace{0101100}}{)}_2=(-1.0101100101)\cdot 2^7$$

[!step] Step 3. Apply the excess bias
Offset exponent $(7{)}_{10}=(111{)}_2$ with excess $(127{)}_{10}=(01111111{)}_2$:

$$\begin{array}{rl}& 01111111\\ +& 00000111\\ & \\ =& 10000110\end{array}$$

Step 4. Assemble the layout

$$\underset{1}{\underbrace{1}}\underset{8}{\underbrace{10000110}}\underset{1}{\underbrace{1}}\underset{23}{\underbrace{01011001010000000000000}}$$

The implicit bit ($1$, red) can be omitted since it will always be $1$. Thus, the layout is:

$$\underset{1}{\underbrace{1}}\underset{8}{\underbrace{10000110}}\underset{23}{\underbrace{01011001010000000000000}}$$

The result is a 32 bit word. The most significant bit $1$ (cyan) signals that the number is negative.