当 64 位 int 在 C/C++ 中转换为 64 位浮点数并且没有完全匹配时，它是否总是落在非小数上？

Question

When int64_t is cast to double and doesn't have an exact match, to my knowledge I get a sort of best-effort-nearest-value equivalent in double. For example, 9223372036854775000 in int64_t appears to end up as 9223372036854774784.0 in double:

#include <stdio.h>

int main(int argc, const char **argv) {
    printf("Corresponding double: %f\n", (double)9223372036854775000LL);
    // Outputs: 9223372036854774784.000000
    return 0;
}

It appears to me as if an int64_t cast to a double always ends up on as a clean non-fractional number, even in this higher number range where double has really low precision. However, I just observed this from random attempts. Is this guaranteed to happen for any value of int64_t cast to a double?

And if I cast this non-fractional double back to int64_t, will I always get the exact corresponding 64bit int with the.0 chopped off? ( Assuming it doesn't overflow during the conversion back. ) Like here:

#include <inttypes.h>
#include <stdio.h>

int main(int argc, const char **argv) {
    printf("Corresponding double: %f\n", (double)9223372036854775000LL);
    // Outputs: 9223372036854774784.000000
    printf("Corresponding int to corresponding double: %" PRId64 "\n",
           (int64_t)((double)9223372036854775000LL));
    // Outputs: 9223372036854774784
    return 0;
}

Or can it be imprecise and get me the "wrong" int in some corner cases?

Intuitively and from my tests the answer to both points appears to be "yes", but if somebody with a good formal understanding of the floating point standards and the maths behind it could confirm this that would be really helpful to me. I would also be curious if any known more aggressive optimizations like gcc's -Ofast are known to break any of this.

Answer 1

In general case yes, both should be true. The floating point base needs to be - if not 2, then at least integer and given that, an integer converted to nearest floating point value can never produce non-zero fractions - either the precision suffices or the lowest-order integer digits in the base of the floating type would be zeroed. For example in your case your system uses ISO/IEC/IEEE 60559 binary floating point numbers. When inspected in base 2, it can be seen that the trailing digits of the value are indeed zeroed:

>>> bin(9223372036854775000)
'0b111111111111111111111111111111111111111111111111111110011011000'
>>> bin(9223372036854774784)
'0b111111111111111111111111111111111111111111111111111110000000000'

The conversion of a double without fractions to an integer type, given that the value of the double falls within the range of the integer type should be exact...

Though you still might encounter a quality-of-implementation issue, or an outright bug - for example MSVC currently has a compiler bug where a round-trip conversion of unsigned 32-bit value with MSB set (or just double value between 2³¹ and 2³²-1 converted to unsigned int ) would "overflow" in the conversion and always result in exactly 2³¹.

Answer 2

The following assumes the value being converted is positive. The behavior of negative numbers is analogous.

C 2018 6.3.1.4 2 specifies conversions from integer to real and says:

… If the value being converted is in the range of values that can be represented but cannot be represented exactly, the result is either the nearest higher or nearest lower representable value, chosen in an implementation-defined manner.

This tells us that some integer value x being converted to floating-point can produce a non-integer only if one of the two representable values bounding x is not an integer and x is not representable.

5.2.4.2.2 specifies the model used for floating-point numbers. Each finite floating-point number is represented by a sequence of digits in a certain base b scaled by b ^e for some exponent e . ( b is an integer greater than 1.) Then, if one of the two values bounding x , say p is not an integer, the scaling must be such that the lowest digit in that floating-point number represents a fraction. But if this is the case, then setting all of the digits in p that represent fractions to 0 must produce a new floating-point number that is an integer. If x < p , this integer must be x , and therefore x is representable in the floating-point format. On the other hand, if p < x , we can add enough to each digit that represents a fraction to make it 0 (and produce a carry to the next higher digit). This will also produce an integer representable in the floating-point type ¹ , and it must be x .

Therefore, if conversion of an integer x to the floating-point type would produce a non-integer, x must be representable in the type. But then conversion to the floating-point type must produce x . So it is never possible to produce a non-integer.

Footnote

¹ It is possible this will carry out of all the digits, as when applying it to a three-digit decimal number 9.99, which produces 10.00. In this case, the value produced is the next power of b , if it is in range of the floating-point format. If it is not, the C standard does not define the behavior. Also note the C standard sets minimum requirements on the range that floating-point formats must support which preclude any format from not being able to represent 1, which avoids a degenerate case in which a conversion could produce a number like.999 because it was the largest representable finite value.

Answer 3

When a 64bit int is cast to 64bit float... and doesn't have an exact match, will it always land on a non-fractional number?
Is this guaranteed to happen for any value of int64_t cast to a double ?

For common double : Yes, it always land on a non-fractional number

When there is no match, the result is the closest floating point representable value above or below, depending on rounding mode. Given the characteristics of common double , these 2 bounding values are also whole numbers. When the value is not representable, there is first a nearby whole number one.

---

... if I cast this non-fractional double back to int64_t , will I always get the exact corresponding 64bit int with the.0 chopped off?

No. Edge cases near INT64_MAX fail as the converted value could become a FP value above INT64_MAX . Then conversion back to the integer type incurs: "the new type is signed and the value cannot be represented in it; either the result is implementation-defined or an implementation-defined signal is raised." C17dr § 6.3.1.3 3

#include <limits.h>
#include <string.h>

int main() {
  long long imaxm1 = LLONG_MAX - 1;
  double max = (double) imaxm1;
  printf("%lld\n%f\n", imaxm1, max);
  long long imax = (long long) max;
  printf("%lld\n", imax);
}

9223372036854775806
9223372036854775808.000000
9223372036854775807  // Value here is implementation defined.

---

Deeper exceptions

(Question variation) When an N bit integer type is cast to a floating point and doesn't have an exact match, will it always land on a non-fractional number?

Integer type range exceeds finite float point

Conversion to infinity: With common float , and uint128_t , UINT128_MAX converts to infinity . This is readily possible with extra wide integer types.

int main() {
  unsigned __int128  imaxm1 = 0xFFFFFFFFFFFFFFFF;
  imaxm1 <<= 64;
  imaxm1 |= 0xFFFFFFFFFFFFFFFF;
  double fmax = (float) imaxm1;
  double max = (double) imaxm1;
  printf("%llde27\n%f\n%f\n", (long long) (imaxm1/1000000000/1000000000/1000000000), 
    fmax, max);
}

340282366920e27
inf
340282366920938463463374607431768211456.000000

Floating point precession deep more than range

On some unicorn implementation, with very wide FP precision and small range, the largest finite could, in theory, not practice, be a non-whole number. Then with an even wider integer type, the conversion could result in this non-whole number value. I do not see this as a legit concern of OP's.