Single-precision floating-point format

In computing, single precision is a usually binary floating-point computer numbering format that occupies 4 bytes (32 bits in modern computers) in computer memory.

In IEEE 754-2008 the 32-bit base 2 format is officially referred to as binary32. It was called single in IEEE 754-1985. In older computers, some older floating point format of 4 bytes was used.

One of the first programming languages to provide single- and double-precision floating-point data types was Fortran. Before the wide-spread adoption of IEEE 754-1985, the representation and properties of the double float data type depended on the computer manufacturer and computer model.

Single precision binary floating-point is used due to its wider range over fixed point (of the same bit-width), even if at the cost of precision.

Single precision is known as float in C, C++ and Java^[1].

IEEE 754 single precision binary floating-point format: binary32

The IEEE 754 standard specifies a binary32 as having:

Sign bit: 1 bit
Exponent width: 8 bits
Significand precision: 24 (23 explicitly stored)

The true significand includes an implicit leading bit with value 1 unless the exponent is stored with all zeros. Thus only 23 bits of the significand appear in the memory format but the total precision is 24 bits (equivalent to log₁₀(2²⁴) ≈ 7.225 decimal digits). The bits are laid out as follows:

Exponent encoding

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

The stored exponents 00_H and FF_H are interpreted specially.

Exponent	Significand zero	Significand non-zero	Equation
00_H	zero, −0	subnormal numbers	(-1)^signbits×2^-126× 0.significandbits
01_H, ..., FE_H	normalized value		(-1)^signbits×2^{exponentbits-127}× 1.significandbits
FF_H	±infinity	NaN (quiet, signalling)

The minimum positive (subnormal) value is 2⁻¹⁴⁹ ≈ 1.4 × 10⁻⁴⁵. The minimum positive normal value is 2⁻¹²⁶ ≈ 1.18 × 10⁻³⁸. The maximum representable value is (2−2⁻²³) × 2¹²⁷ ≈ 3.4 × 10³⁸.

Single precision examples

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

3f80 0000   = 1
c000 0000   = −2

7f7f ffff   ≈ 3.4028234 × 10³⁸  (max single precision)

0000 0000   = 0
8000 0000   = −0

7f80 0000   = infinity
ff80 0000   = -infinity
				
3eaa aaab   ≈ 1/3

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

Converting from single precision binary to human readable form (decimal)

We start with the hexadecimal representation of the value, 41c80000, in this example, and convert it to binary

41c8 0000₁₆ = 0100 0001 1100 1000 0000 0000 0000 0000₂

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

Sign bit: 0
Exponent: 1000 0011₂ = 83₁₆ = 131
Significand: 100 1000 0000 0000 0000 0000₂ = 480000₁₆

We then add the implicit 24th bit to the significand

Significand: 1100 1000 0000 0000 0000 0000₂ = C80000₁₆

and decode the exponent value by subtracting 127

Raw exponent: 83₁₆ = 131
Decoded exponent: 131 - 127 = 4

Each of the 24 bits of the significand, 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  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: 1 + 0.5 + 0.0625 = 1.5625 = C80000/2²³

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

1.5625 × 2⁴ = 25

Thus

41c8 0000   = 25

This is equivalent to:

 $n=(-1)^{s}\times (1+m2^{-23})\times 2^{x-127}$

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

External links

References

^ http://java.sun.com/docs/books/tutorial/java/nutsandbolts/datatypes.html

[1] ttp://java.sun.com/docs/books/tutorial/java/nutsandbolts/datatypes.html

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