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

types floating-point ieee-754

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.

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".

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.

possible duplicate of What's the first double that deviates from its corresponding long by delta?

duplicate of stackoverflow.com/questions/1848700/… ?

@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…”

Dan Bechard

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

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.

Ok, thanks. I got confused, sorry about that. I have another question though: after 16777216, shouldn't the next integer that is representable be 2*16777216? When I run a similar program, I get 16777218 by adding 2 to 16777126.

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.

thus spake a.k.

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.

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

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

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