Bit wise operation of float64

[alex_He]

Hi
I came across an article of how to find the value of 1/sqrt(x) quick.
the idea is to use the log bit wise and pointer operation ( floating point representation with the exponents and mantissas)
ie:
var y float64
var i int64
i= *(long *)&y //evil floating point bit hack

I do not quite understand this operation.
when I try to run it on goplay gound,
an error occur:

func main() {
var y = 2.15
var i int64
i=*(int64 *)&y // error syntax error: unexpected ), expecting expression
i= 0x5f3759df - (i>>1)
y= (float64)&i //error syntax error: unexpected ), expecting expression
fmt.Println(y)
}
could someone please enlighten me ?
Thanks

[petrus]

// Fast inverse square root
// https://en.wikipedia.org/wiki/Fast_inverse_square_root

package main

import (
	"fmt"
	"math"
)

func RSqrt(number float64) float64 {
	const threehalfs = 1.5

	x2 := number * 0.5
	y := number
	i := int64(math.Float64bits(y))   // evil floating point bit level hacking
	i = 0x5FE6EB50C7B537A9 - (i >> 1) // what the f@#!?
	y = math.Float64frombits(uint64(i))
	y = y * (threehalfs - (x2 * y * y)) // 1st iteration
	y = y * (threehalfs - (x2 * y * y)) // 2nd iteration
	y = y * (threehalfs - (x2 * y * y)) // 3rd iteration
	y = y * (threehalfs - (x2 * y * y)) // 4th iteration

	return y
}

func main() {
	number := 0.15625
	fmt.Printf("%g\n", number)

	rsqrt := RSqrt(number)
	fmt.Printf("%g\n", rsqrt)

	rsqrt = 1 / math.Sqrt(number)
	fmt.Printf("%g\n", rsqrt)
}

https://play.golang.org/p/o5w-TaqKjZQ

0.15625
2.5298221281347035
2.5298221281347035

[alex_He]

Hi Petrus
Thank you very much for your help.
More importantly, I learn from you how to do the evil floating point bit level in golang.
once again Thank you very much!