InvSqrt()函数结果不对，求解

[icesongqiang]

#include <iostream>
#include <cmath>
#include <ctime>
#include <windows.h>
void sqrt_inv(double x)
{
	cout << "x= " << x << endl;
	clock_t start = clock();
	const float threehalfs = 1.5F;
	double xhalf = 0.5f*x;
	int i = *(int*)&x;              // get bits for floating VALUE 
	cout << "i= " << i << endl;
	i = 0x5f375a86 - (i >> 1);      // gives initial guess y0
	x = *(float*)&i;                // convert bits BACK to float
	cout << "x= " << x << endl;
	x = x*(threehalfs - xhalf*x*x); // Newton step, repeating increases accuracy
	x = x*(threehalfs - xhalf*x*x); // Newton step, repeating increases accuracy
	x = x*(threehalfs - xhalf*x*x); // Newton step, repeating increases accuracy
	double ans = 1.0 / x;
	Sleep(1000);
	clock_t finish = clock();

	cout << "inv:" << "\t" << ans << "\t" << start << "\t" << finish << "\t" << (finish - start) << endl;  // (finish - start) / CLOCKS_PER_SEC
}
int main()
{
	double x = 2;
	//sqrt_sys(x);
	sqrt_inv(x);
	//sqrt_Newton(x);
	system("pause");
	return 0;
}

输出：

[日立奔腾浪潮微软松下联想]

这个carmack的快速开平方根倒数算法只适用于float，如果double参数的话要： double InvSqrt(double x) { double xhalf = 0.5 * x; long long i = *(long long *)&x; i = 0x5fe6ec85e7de30da - (i >> 1); x = *(double *)&i; x = x * (1.5 - xhalf * x * x); return x; }

[icesongqiang]

void sqrt_inv(float x)     // 参数类型是float
{
	cout << "x= " << x << endl;
	clock_t start = clock();
	const float threehalfs = 1.5F;
	double xhalf = 0.5f*x;
	int i = *(int*)&x;              // get bits for floating VALUE 
	cout << "i= " << i << endl;
	i = 0x5f375a86 - (i >> 1);      // gives initial guess y0
	x = *(float*)&i;                // convert bits BACK to float
	cout << "x= " << x << endl;
	x = x*(threehalfs - xhalf*x*x); // Newton step, repeating increases accuracy
	x = x*(threehalfs - xhalf*x*x); // Newton step, repeating increases accuracy
	x = x*(threehalfs - xhalf*x*x); // Newton step, repeating increases accuracy
	double ans = 1.0 / x;
	Sleep(1000);
	clock_t finish = clock();

	cout << "float_inv:" << "\t" << ans << "\t" << start << "\t" << finish << "\t" << (finish - start) << endl;  // (finish - start) / CLOCKS_PER_SEC
}

void sqrt_inv(double x)     // 类型是double
{
	clock_t start = clock();
	const double threehalfs = 1.5F;
	double xhalf = 0.5f*x;
	long long i = *(long long*)&x;              // get bits for floating VALUE 
	cout << "i= " << i << endl;
	i = 0x5fe6ec85e7de30da - (i >> 1);      // gives initial guess y0
	x = *(double*)&i;                // convert bits BACK to float
	cout << "x= " << x << endl;
	x = x*(threehalfs - xhalf*x*x); // Newton step, repeating increases accuracy
	x = x*(threehalfs - xhalf*x*x); // Newton step, repeating increases accuracy
	x = x*(threehalfs - xhalf*x*x); // Newton step, repeating increases accuracy
	double ans = 1.0 / x;
	Sleep(1000);
	clock_t finish = clock();

	cout << "double_inv:" << "\t" << ans << "\t" << start << "\t" << finish << "\t" << (finish - start) << endl;  // (finish - start) / CLOCKS_PER_SEC
}

[日立奔腾浪潮微软松下联想]

其实迭代一次就可以了，迭代三次精度也没提高多少，性能下降不少。要快的话可以使用rsqrtss指令，速度很好，精度一般，对于VC++、intel C++可以直接使用_mm_rsqrt_ss，不需要内嵌汇编。