A fast method to round a double to a 32-bit int explained

Question

When reading Lua’s source code, I noticed that Lua uses a macro to round double values to 32-bit int values. The macro is defined in the Llimits.h header file and reads as follows:

union i_cast {double d; int i[2]};
#define double2int(i, d, t) \
    {volatile union i_cast u; u.d = (d) + 6755399441055744.0; \
    (i) = (t)u.i[ENDIANLOC];}

Here ENDIANLOC is defined according to endianness: 0 for little endian, 1 for big endian architectures; Lua carefully handles endianness. The t argument is substituted with an integer type like int or unsigned int.

I did a little research and found that there is a simpler format of that macro which uses the same technique:

#define double2int(i, d) \
    {double t = ((d) + 6755399441055744.0); i = *((int *)(&t));}

Or, in a C++-style:

inline int double2int(double d)
{
    d += 6755399441055744.0;
    return reinterpret_cast<int&>(d);
}

This trick can work on any machine using IEEE 754 (which means pretty much every machine today). It works for both positive and negative numbers, and the rounding follows Banker’s Rule. (This is not surprising, since it follows IEEE 754.)

I wrote a little program to test it:

int main()
{
    double d = -12345678.9;
    int i;
    double2int(i, d)
    printf("%d\n", i);
    return 0;
}

And it outputs -12345679, as expected.

I would like to understand how this tricky macro works in detail. The magic number 6755399441055744.0 is actually 2⁵¹ + 2⁵², or 1.5 × 2⁵², and 1.5 in binary can be represented as 1.1. When any 32-bit integer is added to this magic number—

Well, I’m lost from here. How does this trick work?

Update

1. As @Mysticial points out, this method does not limit itself to a 32-bit int, it can also be expanded to a 64-bit int as long as the number is in the range of 2⁵². (Although the macro needs some modification.)

2. Some materials say this method cannot be used in Direct3D.

3. When working with Microsoft assembler for x86, there is an even faster macro written in assembly code (the following is also extracted from Lua source):

    #define double2int(i,n)  __asm {__asm fld n   __asm fistp i}
   

4. There is a similar magic number for single precision numbers: 1.5 × 2²³.

Answer 1

A value of the double floating-point type is represented like so:

and it can be seen as two 32-bit integers; now, the int taken in all the versions of your code (supposing it’s a 32-bit int) is the one on the right in the figure, so what you are doing in the end is just taking the lowest 32 bits of mantissa.

---

Now, to the magic number; as you correctly stated, 6755399441055744 is 2⁵¹ + 2⁵²; adding such a number forces the double to go into the “sweet range” between 2⁵² and 2⁵³, which, as explained by Wikipedia, has an interesting property:

Between 2⁵² = 4,503,599,627,370,496 and 2⁵³ = 9,007,199,254,740,992, the representable numbers are exactly the integers.

This follows from the fact that the mantissa is 52 bits wide.

The other interesting fact about adding 2⁵¹ + 2⁵² is that it affects the mantissa only in the two highest bits—which are discarded anyway, since we are taking only its lowest 32 bits.

---

Last but not least: the sign.

IEEE 754 floating point uses a magnitude and sign representation, while integers on “normal” machines use 2’s complement arithmetic; how is this handled here?

We talked only about positive integers; now suppose we are dealing with a negative number in the range representable by a 32-bit int, so less (in absolute value) than (−2³¹ + 1); call it −a. Such a number is obviously made positive by adding the magic number, and the resulting value is 2⁵² + 2⁵¹ + (−a).

Now, what do we get if we interpret the mantissa in 2’s complement representation? It must be the result of 2’s complement sum of (2⁵² + 2⁵¹) and (−a). Again, the first term affects only the upper two bits, what remains in the bits 0–50 is the 2’s complement representation of (−a) (again, minus the upper two bits).

Since reduction of a 2’s complement number to a smaller width is done just by cutting away the extra bits on the left, taking the lower 32 bits gives us correctly (−a) in 32-bit, 2’s complement arithmetic.

Answer 2

This kind of "trick" comes from older x86 processors, using the 8087 intructions/interface for floating point. On these machines, there's an instruction for converting floating point to integer "fist", but it uses the current fp rounding mode. Unfortunately, the C spec requires that fp->int conversions truncate towards zero, while all other fp operations round to nearest, so doing an
fp->int conversion requires first changing the fp rounding mode, then doing a fist, then restoring the fp rounding mode.

Now on the original 8086/8087, this wasn't too bad, but on later processors that started to get super-scalar and out-of-order execution, altering the fp rounding mode generally serializes the CPU core and is quite expensive. So on a CPU like a Pentium-III or Pentium-IV, this overall cost is quite high -- a normal fp->int conversion is 10x or more expensive than this add+store+load trick.

On x86-64, however, floating point is done with the xmm instructions, and the cost of converting
fp->int is pretty small, so this "optimization" is likely slower than a normal conversion.

Answer 3

if this helps with visualization, that lua magic value

  (2^52+2^51, or base2 of 110 then [50 zeros]

hex

  0x  0018 0000 0000 0000 (18e12)

octal

  0 300 00000 00000 00000 ( 3e17)

Answer 4

Here is a simpler implementation of the above Lua trick:

/**
 * Round to the nearest integer.
 * for tie-breaks: round half to even (bankers' rounding)
 * Only works for inputs in the range: [-2^51, 2^51]
 */
inline double rint(double d)
{
    double x = 6755399441055744.0;  // 2^51 + 2^52
    return d + x - x;
}

The trick works for numbers with absolute value < 2 ^ 51.

This is a little program to test it: ideone.com

#include <cstdio>

int main()
{
    // round to nearest integer
    printf("%.1f, %.1f\n", rint(-12345678.3), rint(-12345678.9));

    // test tie-breaking rule
    printf("%.1f, %.1f, %.1f, %.1f\n", rint(-24.5), rint(-23.5), rint(23.5), rint(24.5));      
    return 0;
}

// output:
// -12345678.0, -12345679.0
// -24.0, -24.0, 24.0, 24.0