求助：将两个C代码转换成Java代码

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【Python Turtle合集】有趣好玩的代码当然要分享给大家啦~（皮卡丘、玫瑰花、小黄人......）前言 🚀 作者：“程序员梨子” 🚀 **文章简介 **：本篇文章主要是写了利用Turtle库绘制四种不一样的图案的小程序！ 🚀 **文章源码免费获取：为了感谢每一个关注我的小可爱💓每篇文章的项目源码都是无偿分享滴💓👇👇👇👇 点这里蓝色这行字体自取，需要什么源码记得说标题名字哈！私信我也可！ 🚀 欢迎小伙伴们点赞👍、收藏⭐、留言💬 正文今天给大家介绍一个好玩儿的库，嘻嘻！ Turtle库是Python语言中一个很流行的绘制图像的函数库，想象一个小乌龟，在一个横轴为x、纵阅读详情

求助，将回归算法的C代码转换成Java代码，VC2010项目代码：http://pan.baidu.com/s/1c14FR0k
主要是两个c文件：multiple_linear_regression_call.c和multiple_linear_regression.c，请哪位大侠改写成Java代码。

multiple_linear_regression_call.c

// multiple_linear_regression_call.cpp : Defines the entry point for the console application.

//



#include "multiple_linear_regression.h"

#include <stdio.h>

#include <math.h>



double data[15][5] = {

	{ 316, 1536, 874, 981, 3894 },

	{ 385, 1771, 777, 1386, 4628 },

	{ 299, 1565, 678, 1672, 4569 },

	{ 326, 1970, 785, 1864, 5340 },

	{ 441, 1890, 785, 2143, 5449 },

	{ 460, 2050, 709, 2176, 5599 },

	{ 470, 1873, 673, 1769, 5010 },

	{ 504, 1955, 793, 2207, 5694 },

	{ 348, 2016, 968, 2251, 5792 },

	{ 400, 2199, 944, 2390, 6126 },

	{ 496, 1328, 749, 2287, 5025 },

	{ 497, 1920, 952, 2388, 5924 },

	{ 533, 1400, 1452, 2093, 5657 },

	{ 506, 1612, 1587, 2083, 6019 },

	{ 458, 1613, 1485, 2390, 6141 },

};



void display(double *dat, double *answer, double *square_poor, int rows, int cols)

{

	double v, *p;

	int i, j;

	printf("regression equation;   Y = %.5lf", answer[0]);

	for (i = 1; i < cols; i ++)

		printf(" + %.5lf*X%d", answer[i], i);

	printf(" ");

	printf("回归显著性检验: ");

	printf("回归平方和：%12.4lf  回归方差：%12.4lf ", square_poor[0], square_poor[2]);

	printf("剩余平方和：%12.4lf  剩余方差：%12.4lf ", square_poor[1], square_poor[3]);

	printf("离差平方和：%12.4lf  标准误差：%12.4lf ", square_poor[0] + square_poor[1], sqrt(square_poor[3]));

	printf("F   检  验：%12.4lf  相关系数：%12.4lf ", square_poor[2] / square_poor[3],

		sqrt(square_poor[0] / (square_poor[0] + square_poor[1])));

	printf("剩余分析: ");

	printf("      观察值      估计值      剩余值    剩余平方 ");

	for (i = 0, p = dat; i < rows; i ++, p ++)

	{

		v = answer[0];

		for (j = 1; j < cols; j ++, p ++)

			v += *p * answer[j];

		printf("%12.2lf%12.2lf%12.2lf%12.2lf ", *p, v, *p - v, (*p - v) * (*p - v));

	}

}



int main(int argc, char* argv[])

{

	double answer[5], square_poor[4];

	/*

	* Y = B0 + B1X1 + B2X2 + ...B4X4

	*/

	if (multiple_linear_regression((double*)data, 15, 5, answer, square_poor) == 0)

		display((double*)data, answer, square_poor, 15, 5);

	system("pause");

	return 0;

}

multiple_linear_regression.c

// MultiLinearRegression.cpp : Defines the exported functions for the DLL application.

//

#include <stdlib.h>

#include <string.h>

#include "multiple_linear_regression.h"



void free_data(double **dat, double *d, int count)

{

	int i;

	free(d);

	for (i = 0; i < count; i ++)

		free(dat[i]);

	free(dat);

}



/*

 * data[count*(count+1)] matrix array 

 * count：equation element

 * answer[count]:answer array

 * return:0--sucess，-1--no solution or infinite solution

*/

int linear_equations(double *data, int count, double *answer)

{

	int j, m, n;

	double tmp, **dat, *d = data;

	dat = (double**)malloc(count * sizeof(double*));

	for (m = 0; m < count; m ++, d += (count + 1))

	{

		dat[m] = (double*)malloc((count + 1) * sizeof(double));

		memcpy(dat[m], d, (count + 1) * sizeof(double));

	}

	d = (double*)malloc((count + 1) * sizeof(double));

	for (m = 0; m < count - 1; m ++)

	{

		for (n = m + 1; n < count && dat[m][m] == 0.0; n ++)

		{

			if ( dat[n][m] != 0.0)

			{

				memcpy(d, dat[m], (count + 1) * sizeof(double));

				memcpy(dat[m], dat[n], (count + 1) * sizeof(double));

				memcpy(dat[n], d, (count + 1) * sizeof(double));

			}

		}

		if (dat[m][m] == 0.0)

		{

			free_data(dat, d, count);

			return -1;

		}

		for (n = m + 1; n < count; n ++)

		{

			tmp = dat[n][m] / dat[m][m];

			for (j = m; j <= count; j ++)

				dat[n][j] -= tmp * dat[m][j];

		}

	}

	for (j = 0; j < count; j ++)

		d[j] = 0.0;

	answer[count - 1] = dat[count - 1][count] / dat[count - 1][count - 1];

	for (m = count - 2; m >= 0; m --)

	{

		for (j = count - 1; j > m; j --)

			d[m] += answer[j] * dat[m][j];

		answer[m] = (dat[m][count] - d[m]) / dat[m][m];

	}

	free_data(dat, d, count);

	return 0;

}





int32_t

multiple_linear_regression(double *data,

	int32_t rows,

	int32_t columns,

	double *answer,

	double *square_poor)

{

	int m, n, i, count = columns - 1;

	double *dat, *p, a, b;

	if (data == 0 || answer == 0 || rows < 2 || columns < 2)

		return -1;

	dat = (double*)malloc(columns * (columns + 1) * sizeof(double));

	dat[0] = (double)rows;

	for (n = 0; n < count; n ++)                     // n = 0 to columns - 2

	{

		a = b = 0.0;

		for (p = data + n, m = 0; m < rows; m ++, p += columns)

		{

			a += *p;

			b += (*p * *p);

		}

		dat[n + 1] = a;                              // dat[0, n+1] = Sum(Xn)

		dat[(n + 1) * (columns + 1)] = a;               // dat[n+1, 0] = Sum(Xn)

		dat[(n + 1) * (columns + 1) + n + 1] = b;       // dat[n+1,n+1] = Sum(Xn * Xn)

		for (i = n + 1; i < count; i ++)             // i = n+1 to columns - 2

		{

			for (a = 0.0, p = data, m = 0; m < rows; m ++, p += columns)

				a += (p[n] * p[i]);

			dat[(n + 1) * (columns + 1) + i + 1] = a;   // dat[n+1, i+1] = Sum(Xn * Xi)

			dat[(i + 1) * (columns + 1) + n + 1] = a;   // dat[i+1, n+1] = Sum(Xn * Xi)

		}

	}

	for (b = 0.0, m = 0, p = data + n; m < rows; m ++, p += columns)

		b += *p;

	dat[columns] = b;                                   // dat[0, columns] = Sum(Y)

	for (n = 0; n < count; n ++)

	{

		for (a = 0.0, p = data, m = 0; m < rows; m ++, p += columns)

			a += (p[n] * p[count]);

		dat[(n + 1) * (columns + 1) + columns] = a;        // dat[n+1, columns] = Sum(Xn * Y)

	}

	n = linear_equations(dat, columns, answer);         

	if (n == 0 && square_poor)

	{

		b = b / rows;                                

		square_poor[0] = square_poor[1] = 0.0;

		p = data;

		for (m = 0; m < rows; m ++, p ++)

		{

			for (i = 1, a = answer[0]; i < columns; i ++, p ++)

				a += (*p * answer[i]);               

			square_poor[0] += ((a - b) * (a - b));    

			square_poor[1] += ((*p - a) * (*p - a));  

		}

		square_poor[2] = square_poor[0] / count;       

		if (rows - columns > 0.0)

			square_poor[3] = square_poor[1] / (rows - columns); 

		else

			square_poor[3] = 0.0;

	}

	free(dat);

	return n;

	return 0;

}

...全文

嵌入式状态机编程-QP状态机框架与常见状态机方法状态机基本术语现态：是指当前所处的状态。条件：又称为“事件”，当一个条件被满足，将会触发一个动作，或者执行一次状态的迁移。动作：条件满足后执行的动作。动作执行完毕后，可以迁移到新的状态，也可以仍旧保持原状态。动作不是必需的，当条件满足后，也可以不执行任何动作，直接迁移到新状态。次态：条件满足后要迁往的新状态。“次态”是相对于“现态”而言的，“次态”一旦被激活，就转变成新的“现态”了。传统有限状态机Fsm实现方法如图，是一个定时计数器，计数器存在两种状态，一种为设置状态，一种为计时状态设阅读详情

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class multiple_linear_regression
{
	// java不需要free_data
	// java没有指针，故使用类属性与方法实现输入与输出
	private double[] data;
	private int rows;
	private int columns;
	private double[] answer;
	private double[] square_poor;
	private int ret;
	public double[] getAnswer()
	{
		return this.answer;	
	}
	public double[] getSquare_poor()
	{
		return this.square_poor;
	}
	public int getRet()
	{
		return this.ret;
	}
	private int linear_equations(double[] data)
	{
		int count = this.columns;
		this.answer = new double[count];

		double dat[][] = new double[count][];
		int di = 0; // d换成索引di
		for (int m = 0; m != count; ++m, di += (count + 1))
		{
			dat[m] = new double[count + 1];
			System.arraycopy(data, di, dat[m], 0, count + 1);
		}
		double d[] = new double[count + 1];
		for (int m = 0; m != count - 1; ++m)
		{
			for (int n = m + 1; n < count && dat[m][n] == 0.0; ++n)
			{
				if (dat[n][m] != 0.0)
				{
					d = dat[m];
					dat[m] = dat[n];
					dat[n] = d;
				}
			}
			if (dat[m][m] == 0.0) { return -1; }
			for (int n = m + 1; n < count; ++n)
			{
				double tmp = dat[n][m] / dat[m][m];
				for (int j = m; j <= count; ++j)
				{
					dat[n][j] -= tmp * dat[m][j];
				}
			}		
		}
		for (int j = 0; j != count; ++j)
		{
			d[j] = 0.0;
		}
		this.answer[count - 1] = dat[count - 1][count] / dat[count - 1][count - 1];
		for (int m = count - 2; m >= 0; --m)
		{
			for (int j = count - 1; j > m; --j)
			{
				d[m] += answer[j] * dat[m][j];
			}
			answer[m] = (dat[m][count] - d[m]) / dat[m][m];
		}
		return 0;
	}
	public multiple_linear_regression(double[][] data)
	{
		if (data != null || data.length < 2 || data[0].length < 2)
		{
			this.ret = -1;
		}
		this.rows = data.length;
		this.columns = data[0].length;
		this.data = new double[this.rows * this.columns];
		for (int i = 0; i != this.rows; ++i)
		{
			System.arraycopy(
				data[i], 0, 
				this.data, i * this.columns, 
				this.columns
			);
		}
		
		int count = this.columns - 1;
		double dat[] = new double[this.columns * (this.columns + 1)];
		dat[0] = (double)this.rows;
		int n;
		for (n = 0; n != count; ++n)
		{
			double a = 0.0, b = 0.0;
			for (int pi = n, m = 0; m != rows; ++m, pi += columns) // p换成索引pi
			{
				a += this.data[pi];
				b += this.data[pi] * this.data[pi];
			}
			dat[n + 1] = a;
			dat[(n + 1) * (this.columns + 1)] = a;
			dat[(n + 1) * (this.columns + 1) + n + 1] = b;
			for (int i = n + 1; i < count; ++i)
			{
				a = 0.0;
				for (int pi = 0, m = 0; m != this.rows; ++m, pi += this.columns)
				{
					a += this.data[pi + n] * this.data[pi + i];
				}
				dat[(n + 1) * (this.columns + 1) + i + 1] = a;
				dat[(i + 1) * (this.columns + 1) + n + 1] = a;
			}
		}
		double b = 0.0;
		for (int m = 0, pi = n; m != this.rows; ++m, pi += this.columns)
		{
			b += this.data[pi];
		}
		dat[this.columns] = b;
		for (n = 0; n != count; ++n)
		{	
			double a = 0.0;
			for (int pi = 0, m = 0; m != this.rows; ++m, pi += this.columns)
			{
				a += this.data[pi + n] * this.data[pi + count];
			}
			dat[(n + 1) * (this.columns + 1) + this.columns] = a;
		}
		n = this.linear_equations(dat);
		if (n == 0)
		{
			this.square_poor = new double[4];
			b = b / this.rows;
			this.square_poor[0] = this.square_poor[1] = 0.0;
			for (int m = 0, pi = 0; m < this.rows; ++m, ++pi) // p用pi表示
			{
				double a = this.answer[0];
				for (int i = 1; i < this.columns; ++i, ++pi)
				{
					a += this.data[pi] * this.answer[i];
				}
				this.square_poor[0] += (a - b) * (a - b);
				this.square_poor[1] += (this.data[pi] - a) * (this.data[pi] - a);
			}
			this.square_poor[2] = this.square_poor[0] / count;
			this.square_poor[3] = (
				this.rows - this.columns > 0.0 ?
				this.square_poor[1] / (this.rows - this.columns)
				: 0.0
			);
		}
		this.ret = n;	
	}
	public void display()
	{
		System.out.printf("regression equation;	Y = %.5f" , this.answer[0]);
		for (int i = 1; i < this.columns; ++i)
		{
			System.out.printf(" + %.5f*X%d", this.answer[i], i);
		}
		System.out.print(" ");
		System.out.print("回归显著性检验：");
		System.out.printf("回归平方和: %12.4f  回归方差: %12.4f ", this.square_poor[0], this.square_poor[2]);
		System.out.printf("剩余平方和: %12.4f  剩余方差: %12.4f ", this.square_poor[1], this.square_poor[3]);
		System.out.printf("离差平方和: %12.4f  标准误差: %12.4f ", this.square_poor[0] + this.square_poor[1], Math.sqrt(this.square_poor[3]));
		System.out.printf("F   检  验: %12.4f  相关系数: %12.4f ", this.square_poor[2] / this.square_poor[3],
			Math.sqrt(this.square_poor[0] / (this.square_poor[0] + this.square_poor[1])));
		System.out.print("剩余分析: ");
		System.out.print("      观察值      估计值      剩余值     剩余平方 ");
		double v;
		for (int i = 0, pi = 0; i != this.rows; ++i, ++pi)
		{
			v = this.answer[0];
			for (int j = 1; j < this.columns; ++j, ++pi)
			{
				v += this.data[pi] * this.answer[j];
			}
			System.out.printf("%12.2f%12.2f%12.2f%12.2f ", this.data[pi], v, this.data[pi] - v, (this.data[pi] - v) * (this.data[pi] - v));
		}
	}
	public static void main(String[] args)
	{
		double data[][] = {
			{ 316, 1536, 874, 981, 3894 },
			{ 385, 1771, 777, 1386, 4628 },
			{ 299, 1565, 678, 1672, 4569 },
			{ 326, 1970, 785, 1864, 5340 },
			{ 441, 1890, 785, 2143, 5449 },
			{ 460, 2050, 709, 2176, 5599 },
			{ 470, 1873, 673, 1769, 5010 },
			{ 504, 1955, 793, 2207, 5694 },
			{ 348, 2016, 968, 2251, 5792 },
			{ 400, 2199, 944, 2390, 6126 },
			{ 496, 1328, 749, 2287, 5025 },
			{ 497, 1920, 952, 2388, 5924 },
			{ 533, 1400, 1452, 2093, 5657 },
			{ 506, 1612, 1587, 2083, 6019 },
			{ 458, 1613, 1485, 2390, 6141 }
		};
		multiple_linear_regression m = new multiple_linear_regression(data);
		if (m.getRet() == 0)
		{
			m.display();
		}
	}
}

不知道对不对. 基本照抄, 只是把malloc换成new, 指针换成索引.

求助：将两个C代码转换成Java代码

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