Question

I’m looking to for a reasonably efficient way of determining if a floating point value (double) can be exactly represented by an integer data type (long, 64 bit).

My initial thought was to check the exponent to see if it was 0 (or more precisely 127). But that won’t work because 2.0 would be e=1 m=1…

So basically, I am stuck. I have a feeling that I can do this with bit masks, but I’m just not getting my head around how to do that at this point.

So how can I check to see if a double is exactly representable as a long?

Thanks

Answer 1

Here’s one method that could work in most cases. I’m not sure if/how it will break if you give it NaN, INF, very large (overflow) numbers…

(Though I think they will all return false – not exactly representable.)

You could:

1. Cast it to an integer.
2. Cast it back to a floating-point.
3. Compare with original value.

Something like this:

double val = ... ;  //  Value

if ((double)(long long)val == val){
    //  Exactly representable
}

floor() and ceil() are also fair game (though they may fail if the value overflows an integer):

floor(val) == val
ceil(val) == val

---

And here’s a messy bit-mask solution:
This uses union type-punning and assumes IEEE double-precision. Union type-punning is only valid in C99 TR2 and later.

int representable(double x){
    //  Handle corner cases:
    if (x == 0)
      return 1;

    //  -2^63 is representable as a signed 64-bit integer, but +2^63 is not.
    if (x == -9223372036854775808.)
      return 1;

    //  Warning: Union type-punning is only valid in C99 TR2 or later.
    union{
        double f;
        uint64_t i;
    } val;

    val.f = x;

    uint64_t exp = val.i & 0x7ff0000000000000ull;
    uint64_t man = val.i & 0x000fffffffffffffull;
    man |= 0x0010000000000000ull;  //  Implicit leading 1-bit.

    int shift = (exp >> 52) - 1075;
    //  Out of range
    if (shift < -52 || shift > 10)
        return 0;

    //  Test mantissa
    if (shift < 0){
        shift = -shift;
        return ((man >> shift) << shift) == man;
    }else{
        return ((man << shift) >> shift) == man;
    }
}