# How to convert from floating point binary to decimal in half precision(16 bits)?

I'm trying to convert a 16 bit precision binary number to decimal format however I am completely failing to do so.

The binary I'm trying to convert is $0101011101010000$ My current method is:

Separation: $0|10101|1101010000$

Sign = 0

Mantissa = $1.1101010000$

Exponent = $21 - (2^4 - 1) = 6$

Mantissa Denormalised = $1110101.0000$

This gives an answer of 117. Is this actually correct or am I making a mistake in my method?

• It seems to be correct ! – Xoff Jan 31 '15 at 22:00
• Why do you say "I am completely failing to do so"? – TonyK Jan 31 '15 at 22:16
• Ah, I was using this binary example to check whether my method was correct as it was not working for a different number. All that happened was that I failed in moving the decimal point correctly... – echoeida Jan 31 '15 at 22:27

## 2 Answers

You are right.

You can do that automatically with python and numpy :

import numpy as np
import struct
a=struct.pack("H",int("0101011101010000",2))
np.frombuffer(a, dtype =np.float16)


and you get : 117.0

Your formula produces the correct result 117.0 in this case but it may fail for subnormal numbers, for NaNs, for +/- infinity.

>>> float_from_unsigned16(int("0101011101010000", 2))
117.0


where float_from_unsigned16(n) (in Python):

def float_from_unsigned16(n):
assert 0 <= n < 2**16
sign = n >> 15
exp = (n >> 10) & 0b011111
fraction = n & (2**10 - 1)
if exp == 0:
if fraction == 0:
return -0.0 if sign else 0.0
else:
return (-1)**sign * fraction / 2**10 * 2**(-14)  # subnormal
elif exp == 0b11111:
if fraction == 0:
return float('-inf') if sign else float('inf')
else:
return float('nan')
return (-1)**sign * (1 + fraction / 2**10) * 2**(exp - 15)