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

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.

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.

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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.

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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.