# Bit patterns of float

I’ve spent too many hours repeatedly trying to find this information on the Web. Time to write it down.

On the vast majority of computers, float is 32 bits, and it follows the IEEE 754 standard (also known historically as IEC 559). If you look at reinterpret_cast<int&>(myfloat) — or, in C++20, std::bit_cast<int>(myfloat) — you’ll find that the bits go in this order:

Sign Biased exponent Mantissa
1 8 23

For example, 314.0f reinterprets into int(0x439d0000):

 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

The biased exponent bits are 10000111, or 127+8. The mantissa bits are 00111010000000000000000. Stick the implicit leading 1 on the front and we get $$1.0011101_2\times 2^8$$, or $$100111010_2$$, which is binary for 314.

Notice that this is independent of byte endianness, as long as your computer uses the same endianness for both ints and floats. If you treat 314.0f as an array of bytes, you might find that it’s 43 9d 00 00 on a big-endian machine and 00 00 9d 43 on a little-endian machine; but the float’s sign bit will always correspond to the int’s sign bit, and so on.

Now for the parts I always have trouble finding on the Web.

Zero is all-bits-zero. Its biased exponent is 00000000, or 127−127. So, if it were interpreted as an ordinary number, its value would be $$1.0_2\times 2^{-127}$$; but it’s not. It’s just “zero.”

 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Any non-zero number with an all-bits-zero biased exponent is “denormal” or “subnormal”; its mantissa does not have an implicit leading 1 bit, and its effective exponent “sticks” at $$2^{-126}$$. Here are three denormals. The first one’s value is approximately 1.40e-45. The middle one’s value is $$0.1_2\times 2^{-126}$$, or approximately 5.88e-39. The third’s value is just a hair less than FLT_MIN.

 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

FLT_MIN, a.k.a. std::numeric_limits<float>::min(), is approximately 1.18e-38. Its biased exponent is 00000001, or 127−126, and its mantissa is all-bits-zero, so its value is $$1.0_2\times 2^{-126}$$.

 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Next come all the “normal” numbers. For example, the value 1.0 is represented as $$1.0_2\times 2^{0}$$, for a biased exponent of 127+0 and a bit pattern of 3f800000:

 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

and 2.0 is represented as $$1.0_2\times 2^{1}$$, for a biased exponent of 127+1 and a bit pattern of 40000000:

 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

FLT_MAX, a.k.a. std::numeric_limits<float>::max(), is approximately 3.4e+38. Its biased exponent is 11111110, or 127+127.

 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

HUGE_VALF, a.k.a. std::numeric_limits<float>::infinity(), looks like this. Its biased exponent is all-bits-one, and its mantissa is all-bits-zero.

 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

The following three bit-patterns are all signaling NaNs. std::numeric_limits<float>::signaling_NaN() is the middle one. A signaling NaN’s biased exponent is all-bits-one and its mantissa’s top bit is 0. The remaining 22 mantissa bits are “payload.” They can be anything except all-bits-zero (because if the mantissa were all-bits-zero, it’d be HUGE_VALF instead).

 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

 0 1 1 1 1 1 1 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

The following two bit-patterns are both quiet NaNs. NAN, a.k.a. std::numeric_limits<float>::quiet_NaN(), is the first one. A quiet NaN’s biased exponent is all-bits-one and its mantissa’s top bit is 1. The remaining 22 mantissa bits are “payload.” They can be anything.

Observe that every signaling NaN has a “corresponding” quiet NaN: just flip the top bit of its mantissa from 0 to 1. However, there is exactly one quiet NaN which does not correspond to any signaling NaN: flipping the top mantissa bit of NAN gives you HUGE_VAL.

 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Flip the sign bit on any of these bit-patterns and you get negative versions of all the preceding floats.

Negative zero:

 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Negative denormals:

 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

-FLT_MIN, a.k.a. -std::numeric_limits<float>::min(), approximately -1.18e-38:

 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

-FLT_MAX, a.k.a. std::numeric_limits<float>::lowest(), approximately -3.4e+38.

 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

-HUGE_VALF, a.k.a. -std::numeric_limits<float>::infinity():

 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Signaling NaNs with negative sign bits:

 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Quiet NaNs with negative sign bits:

 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

## Implementing IEEE 754’s totalOrder

IEEE 754 specifies a totalOrder predicate on floats (standardized as std::strong_order in C++20) which orders the floats like this:

• Negative quiet NaNs, ordered by payload bits.
• Negative signaling NaNs, ordered by payload bits.
• Negative infinity.
• Negative normal and denormal numbers.
• Negative zero.
• Positive zero.
• Positive normal and denormal numbers.
• Positive infinity.
• Positive signaling NaNs, ordered by payload bits.
• Positive quiet NaNs, ordered by payload bits.

According to Stack Overflow this is equivalent to comparing the bit patterns as if they were sign-magnitude integers (note: not ordinary two’s-complement integers)… with the caveat that negative zero should be ordered less-than positive zero, so if the sign bit was set, you should subtract 1 from the two’s-complement representation before comparing. I believe this can be implemented by the following C++20 algorithm:

constexpr std::strong_ordering totalOrder(float x, float y)
{
int rx = std::bit_cast<int>(x);
int ry = std::bit_cast<int>(y);
rx = (rx < 0) ? (INT_MIN - rx - 1) : rx;
ry = (ry < 0) ? (INT_MIN - ry - 1) : ry;
return rx <=> ry;
}

Posted 2021-09-05