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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.

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  • 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". Commented Sep 25, 2010 at 12:46
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    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. Commented Sep 25, 2010 at 12:56
  • @ks1322 Your edit makes the sentence ungrammatical. “however many” is not equivalent to “how many” and is used correctly in the original sentence (whereas “how many” does not fit). See english-test.net/forum/ftopic9565.html or many other Google results if you want to see more examples of that sort of phrase. Commented Dec 14, 2014 at 13:07
  • @PascalCuoq Completely disagree, even after reading your link (which doesn't seem to say anything about contexts in which "however" fits but "how" doesn't). Commented Dec 24, 2016 at 12:00
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    @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…” Commented Dec 26, 2016 at 23:40

2 Answers 2

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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 32-bit floats, 16,777,217 (224 + 1).
  • For 64-bit floats, 9,007,199,254,740,993 (253 + 1).

Here's an example in CPython 3.10, which uses 64-bit floats:

>>> 9007199254740993.0
9007199254740992.0
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13 Comments

I declared a float and set it equal to 16,777,217. But when I printed it using cout it resulted in 16,777,216. I'm using C++. Why can't I get 16,777,217?
@sodiumnitrate Check the question title. 16777217 is the first integer incapable of being represented exactly.
The next integer is indeed 16777218, because 2 now becomes the last significant binary digit.
In C++, that's (1 << std::numeric_limits<float>::digits) + 1, and in C, (1 << FLT_MANT_DIG) + 1. The former is nice because it can be part of a template. Don't add the +1 if you just want the largest representable integer.
You can use this to examine floating point bit representations and find the min/max integer values: h-schmidt.net/FloatConverter/IEEE754.html here is another one for 16, 32, 64 and 128 bit floating point: weitz.de/ieee
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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 224 can fit and consequently the first unrepresentable integer for float is 224+1. As noted above. Again.

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This clearly explained the "extra implicit bit of precision" part of the other. Thanks.

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