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For an arbitrary value 'v' of a floating point type (float/double/long double), does C89 guarantee that the mathematically exact integer result of floor(v) and ceil(v) is a representable value of the type of 'v'?

Does any of the later C or C++ standards guarantee this?

Does IEEE 754 guarantee this?

Stephen Canon
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Kristian Spangsege
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1 Answers1

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This is guaranteed by the construction of IEEE-754 numbers. (To be clear: C does not guarantee IEEE-754, but the following analysis holds for all other floating-point formats with which I am familiar as well; the crucial property is that all sufficiently large numbers in the format are integers).

Recall that a normal IEEE-754 number has the form ±1.xxx...xxx * 2^n, where the width of the significand field (the xxx...xxx part) is defined by the type of the number (23 binary digits for single precision, 52 binary digits for double precision). All such numbers with an exponent (n) within the allowed range are representable.

Assume WLOG that v is positive (if v were negative, we could swap ceil and floor in the following analysis).

Let v have k significant bits, and write v out as a binary fixed point number; there are three possibilities:

Case 1: All significand bits are integral. When we write out v, it looks like this

xxxxxxxxxxxxxxxxxxxxxxxx000000...00000.0

then v is an integer, and so ceil(v) = floor(v) = v, and so both are trivially representable.

Case 2: All significand bits are fractional. When we write out v, it looks like

0.000000...00000xxxxxxxxxxxxxxxxxxxxxxxx

then v is in the range [0,1), and so floor(v) = 0, which is representable, and ceil(v) is either zero or one, both of which are representable.

Case 3: v contains both integral and fractional significand bits:

xxxxxxxxxxxxxx.xxxxxxxxxx

then floor(v) is just:

xxxxxxxxxxxxxx.

because we have thrown away at least one fractional bit, floor(v) has at most k-1 significant bits, and the same exponent as v, so it is representable.

If v is an integer, then ceil(v) = floor(v) = v, so ceil(v) is representable. Otherwise, ceil(v) = floor(v) + 1, and so also has at most k-1 significant bits and is also representable.

Stephen Canon
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    This answers for IEEE 754, but the C standard might be sloppier. Does it permit a case where something like 99.9999 (or 0xff.ffff or binary 11.1111) is the largest representable finite number, so its ceiling, although it is an integer with fewer digits in its significand, is not representable? – Eric Postpischil Sep 26 '12 at 13:57
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    @EricPostpischil: Yes, the C standard allows such a type. That said, none of the common alternative floating-point formats (or even mildly exotic formats like the old IBM or UNIVAC) fail to have the property that all sufficiently large numbers are integers. – Stephen Canon Sep 26 '12 at 14:04
  • I suspect that for C, floor() and ceil() are guaranteed to return integer values even for non-IEEE implementations (is this true?) However, it must then be possible for floor() to return an integer value that is less than the true integer result, and for ceil() to return an integer value that is greater than the true integer result. – Kristian Spangsege Sep 26 '12 at 18:05