Single-precision floating-point format

The single-precision binary floating-point exponent is encoded using an offset-binary representation, with the zero offset being 127; also known as exponent bias in the IEEE 754 standard.

• E_min = 01_H−7F_H = −126
• E_max = FE_H−7F_H = 127
• Exponent bias = 7F_H = 127

Thus, in order to get the true exponent as defined by the offset-binary representation, the offset of 127 has to be subtracted from the stored exponent.

The stored exponents 00_H and FF_H are interpreted specially.

Exponent	fraction = 0	fraction ≠ 0	Equation
00H = 000000002	±zero	subnormal number	${\displaystyle (-1)^{\text{sign}}\times 2^{-126}\times 0.{\text{fraction}}}$
01H, ..., FEH = 000000012, ..., 111111102	normal value		${\displaystyle (-1)^{\text{sign}}\times 2^{{\text{exponent}}-127}\times 1.{\text{fraction}}}$
FFH = 111111112	±infinity	NaN (quiet, signalling)

The minimum positive normal value is ${\displaystyle 2^{-126}\approx 1.18\times 10^{-38}}$ and the minimum positive (subnormal) value is ${\displaystyle 2^{-149}\approx 1.4\times 10^{-45}}$.

In general, refer to the IEEE 754 standard itself for the strict conversion (including the rounding behaviour) of a real number into its equivalent binary32 format.

Here we can show how to convert a base-10 real number into an IEEE 754 binary32 format using the following outline:

• Consider a real number with an integer and a fraction part such as 12.375
• Convert and normalize the integer part into binary
• Convert the fraction part using the following technique as shown here
• Add the two results and adjust them to produce a proper final conversion

Conversion of the fractional part: Consider 0.375, the fractional part of 12.375. To convert it into a binary fraction, multiply the fraction by 2, take the integer part and repeat with the new fraction by 2 until a fraction of zero is found or until the precision limit is reached which is 23 fraction digits for IEEE 754 binary32 format.

${\displaystyle 0.375\times 2=0.750=0+0.750\Rightarrow b_{-1}=0}$, the integer part represents the binary fraction digit. Re-multiply 0.750 by 2 to proceed

${\displaystyle 0.750\times 2=1.500=1+0.500\Rightarrow b_{-2}=1}$

${\displaystyle 0.500\times 2=1.000=1+0.000\Rightarrow b_{-3}=1}$, fraction = 0.011, terminate

We see that ${\displaystyle (0.375)_{10}}$ can be exactly represented in binary as ${\displaystyle (0.011)_{2}}$. Not all decimal fractions can be represented in a finite digit binary fraction. For example, decimal 0.1 cannot be represented in binary exactly, only approximated. Therefore:

${\displaystyle (12.375)_{10}=(12)_{10}+(0.375)_{10}=(1100)_{2}+(0.011)_{2}=(1100.011)_{2}}$

Since IEEE 754 binary32 format requires real values to be represented in ${\displaystyle (1.x_{1}x_{2}...x_{23})_{2}\times 2^{e}}$ format (see Normalized number, Denormalized number), 1100.011 is shifted to the right by 3 digits to become ${\displaystyle (1.100011)_{2}\times 2^{3}}$

Finally we can see that: ${\displaystyle (12.375)_{10}=(1.100011)_{2}\times 2^{3}}$

From which we deduce:

• The exponent is 3 (and in the biased form it is therefore ${\displaystyle (127+3)_{10}=(130)_{10}=(1000\ 0010)_{2}}$
• The fraction is 100011 (looking to the right of the binary point)

From these we can form the resulting 32-bit IEEE 754 binary32 format representation of 12.375:

${\displaystyle (12.375)_{10}=(0\ 10000010\ 10001100000000000000000)_{2}=(41460000)_{16}}$

Note: consider converting 68.123 into IEEE 754 binary32 format: Using the above procedure you expect to get ${\displaystyle ({\text{42883EF9}})_{16}}$ with the last 4 bits being 1001. However, due to the default rounding behaviour of IEEE 754 format, what you get is ${\displaystyle ({\text{42883EFA}})_{16}}$, whose last 4 bits are 1010.

Example 1: Consider decimal 1. We can see that: ${\displaystyle (1)_{10}=(1.0)_{2}\times 2^{0}}$

From which we deduce:

• The exponent is 0 (and in the biased form it is therefore ${\displaystyle (127+0)_{10}=(127)_{10}=(0111\ 1111)_{2}}$
• The fraction is 0 (looking to the right of the binary point in 1.0 is all ${\displaystyle 0=000...0}$)

From these we can form the resulting 32-bit IEEE 754 binary32 format representation of real number 1:

${\displaystyle (1)_{10}=(0\ 01111111\ 00000000000000000000000)_{2}=({\text{3F800000}})_{16}}$

Example 2: Consider a value 0.25. We can see that: ${\displaystyle (0.25)_{10}=(1.0)_{2}\times 2^{-2}}$

From which we deduce:

• The exponent is −2 (and in the biased form it is ${\displaystyle (127+(-2))_{10}=(125)_{10}=(0111\ 1101)_{2}}$)
• The fraction is 0 (looking to the right of binary point in 1.0 is all zeroes)

From these we can form the resulting 32-bit IEEE 754 binary32 format representation of real number 0.25:

${\displaystyle (0.25)_{10}=(0\ 01111101\ 00000000000000000000000)_{2}=({\text{3E800000}})_{16}}$

Example 3: Consider a value of 0.375. We saw that ${\displaystyle 0.375={(0.011)_{2}}={(1.1)_{2}}\times 2^{-2}}$

Hence after determining a representation of 0.375 as ${\displaystyle {(1.1)_{2}}\times 2^{-2}}$ we can proceed as above:

• The exponent is −2 (and in the biased form it is ${\displaystyle (127+(-2))_{10}=(125)_{10}=(0111\ 1101)_{2}}$)
• The fraction is 1 (looking to the right of binary point in 1.1 is a single ${\displaystyle 1=x_{1}}$)

From these we can form the resulting 32-bit IEEE 754 binary32 format representation of real number 0.375:

${\displaystyle (0.375)_{10}=(0\ 01111101\ 10000000000000000000000)_{2}=({\text{3EC00000}})_{16}}$

If the binary32 value, 41C80000 in this example, is in hexadecimal we first convert it to binary:

${\displaystyle {\text{41C8 0000}}_{16}=0100\ 0001\ 1100\ 1000\ 0000\ 0000\ 0000\ 0000_{2}}$

then we break it down into three parts: sign bit, exponent, and significand.

• Sign bit: ${\displaystyle 0_{2}}$
• Exponent: ${\displaystyle 1000\ 0011_{2}=83_{16}=131_{10}}$
• Significand: ${\displaystyle 100\ 1000\ 0000\ 0000\ 0000\ 0000_{2}=480000_{16}}$

We then add the implicit 24th bit to the significand:

• Significand: ${\displaystyle \mathbf {1} 100\ 1000\ 0000\ 0000\ 0000\ 0000_{2}={\text{C80000}}_{16}}$

and decode the exponent value by subtracting 127:

• Raw exponent: ${\displaystyle 83_{16}=131_{10}}$
• Decoded exponent: ${\displaystyle 131-127=4}$

Each of the 24 bits of the significand (including the implicit 24th bit), bit 23 to bit 0, represents a value, starting at 1 and halves for each bit, as follows:

bit 23 = 1
bit 22 = 0.5
bit 21 = 0.25
bit 20 = 0.125
bit 19 = 0.0625
bit 18 = 0.03125
bit 17 = 0.015625
.
.
bit 6 = 0.00000762939453125
bit 5 = 0.000003814697265625
bit 4 = 0.0000019073486328125
bit 3 = 0.00000095367431640625
bit 2 = 0.000000476837158203125
bit 1 = 0.0000002384185791015625
bit 0 = 0.00000011920928955078125

The significand in this example has three bits set: bit 23, bit 22, and bit 19. We can now decode the significand by adding the values represented by these bits.

• Decoded significand: ${\displaystyle 1+0.5+0.0625=1.5625={\text{C80000}}/2^{23}}$

Then we need to multiply with the base, 2, to the power of the exponent, to get the final result:

${\displaystyle 1.5625\times 2^{4}=25}$

Thus

${\displaystyle {\text{41C8 0000}}=25}$

This is equivalent to:

${\displaystyle n=(-1)^{s}\times (1+m*2^{-23})\times 2^{x-127}}$

where s is the sign bit, x is the exponent, and m is the significand.

• Decimals between 1 and 2: fixed interval 2⁻²³ (1+2⁻²³ is the next largest float after 1)
• Decimals between 2 and 4: fixed interval 2⁻²²
• Decimals between 4 and 8: fixed interval 2⁻²¹
• ...
• Decimals between 2ⁿ and 2ⁿ⁺¹: fixed interval 2^n-23
• ...
• Decimals between 2²²=4194304 and 2²³=8388608: fixed interval 2⁻¹=0.5
• Decimals between 2²³=8388608 and 2²⁴=16777216: fixed interval 2⁰=1

• Integers between 0 and 16777216 can be exactly represented (also applies for negative integers between −16777216 and 0)
• Integers between 2²⁴=16777216 and 2²⁵=33554432 round to a multiple of 2 (even number)
• Integers between 2²⁵ and 2²⁶ round to a multiple of 4
• ...
• Integers between 2ⁿ and 2ⁿ⁺¹ round to a multiple of 2^n-23
• ...
• Integers between 2¹²⁷ and 2¹²⁸ round to a multiple of 2¹⁰⁴
• Integers greater than or equal to 2¹²⁸ are rounded to "infinity".

These examples are given in bit representation, in hexadecimal and binary, of the floating-point value. This includes the sign, (biased) exponent, and significand.

0 00000000 000000000000000000000012 = 0000 000116 = 2−126 × 2−23 = 2−149 ≈ 1.4012984643 × 10−45
                                                   (smallest positive subnormal number)

0 00000000 111111111111111111111112 = 007f ffff16 = 2−126 × (1 − 2−23) ≈ 1.1754942107 ×10−38
                                                   (largest subnormal number)

0 00000001 000000000000000000000002 = 0080 000016 = 2−126 ≈ 1.1754943508 × 10−38
                                                   (smallest positive normal number)

0 11111110 111111111111111111111112 = 7f7f ffff16 = 2127 × (2 − 2−23) ≈ 3.4028234664 × 1038
                                                   (largest normal number)

0 01111110 111111111111111111111112 = 3f7f ffff16 = 1 − 2−24 ≈ 0.999999940395355225
                                                   (largest number less than one)

0 01111111 000000000000000000000002 = 3f80 000016 = 1 (one)

0 01111111 000000000000000000000012 = 3f80 000116 = 1 + 2−23 ≈ 1.00000011920928955
                                                   (smallest number larger than one)

1 10000000 000000000000000000000002 = c000 000016 = −2
0 00000000 000000000000000000000002 = 0000 000016 = 0
1 00000000 000000000000000000000002 = 8000 000016 = −0
                                   
0 11111111 000000000000000000000002 = 7f80 000016 = infinity
1 11111111 000000000000000000000002 = ff80 000016 = −infinity
                                   
0 10000000 100100100001111110110112 = 4049 0fdb16 ≈ 3.14159274101257324 ≈ π ( pi )
0 01111101 010101010101010101010112 = 3eaa aaab16 ≈ 0.333333343267440796 ≈ 1/3
                                   
x 11111111 100000000000000000000012 = ffc0 000116 = qNaN (on x86 and ARM processors)
x 11111111 000000000000000000000012 = ff80 000116 = sNaN (on x86 and ARM processors)

By default, 1/3 rounds up, instead of down like double precision, because of the even number of bits in the significand. The bits of 1/3 beyond the rounding point are 1010... which is more than 1/2 of a unit in the last place.

Encodings of qNaN and sNaN are not specified in IEEE 754 and implemented differently on different processors. The x86 family and the ARM family processors use the most significant bit of the significand field to indicate a quiet NaN. The PA-RISC processors use the bit to indicate a signalling NaN.

The design of floating-point format allows various optimisations, resulting from the easy generation of a base-2 logarithm approximation from an integer view of the raw bit pattern. Integer arithmetic and bit-shifting can yield an approximation to reciprocal square root (fast inverse square root), commonly required in computer graphics.