Which is the first integer that an IEEE 754 float is incapable of representing exactly?

185
votes

For clarity, if I'm using a language that implements IEE 754 floats and I declare:

float f0 = 0.f;
float f1 = 1.f;

...and then print them back out, I'll get 0.0000 and 1.0000 - exactly.

But IEEE 754 isn't capable of representing all the numbers along the real line. Close to zero, the 'gaps' are small; as you get further away, the gaps get larger.

So, my question is: for an IEEE 754 float, which is the first (closest to zero) integer which cannot be exactly represented? I'm only really concerned with 32-bit floats for now, although I'll be interested to hear the answer for 64-bit if someone gives it!

I thought this would be as simple as calculating 2bits_of_mantissa and adding 1, where bits_of_mantissa is how many bits the standard exposes. I did this for 32-bit floats on my machine (MSVC++, Win64), and it seemed fine, though.

2
typesfloating-pointieee-754
Why did you add one if you wanted an irrepresentable number? And what number did you use or get? And is this homework? And your question title says "integer" but your question says "float". – msw
Because I figured that maxing the mantissa would give me the highest representable number. 2^22. No, it's a curiosity question. I've always felt guilty putting ints in floats, even when I know that the int in question is always going to be very small. I want to know what the upper limit is. As far as I can tell, the title and question are the same, just phrased differently. – Floomi
possible duplicate of What's the first double that deviates from its corresponding long by delta? – Andrew Mao
duplicate of stackoverflow.com/questions/1848700/… ? – FrankH.
@KyleStrand reverted^2. I don't know why one seemed more correct to me than the other at the time. Now they both seem awkward compared to “… is the number of bits…” – Pascal Cuoq

2 Answers

234
votes

2mantissa bits + 1 + 1

The +1 in the exponent (mantissa bits + 1) is because, if the mantissa contains abcdef... the number it represents is actually 1.abcdef... × 2^e, providing an extra implicit bit of precision.

Therefore, the first integer that cannot be accurately represented and will be rounded is:
For float, 16,777,217 (224 + 1).
For double, 9,007,199,254,740,993 (253 + 1).

>>> 9007199254740993.0
9007199254740992
43
votes

The largest value representable by an n bit integer is 2n-1. As noted above, a float has 24 bits of precision in the significand which would seem to imply that 224 wouldn't fit.

However.

Powers of 2 within the range of the exponent are exactly representable as 1.0×2n, so 224can fit and consequently the first unrepresentable integer for float is 224+1. As noted above. Again.