100分求助高手帮我把以下C#代码转换成C语言的代码

dopsop110 2010-05-23 08:32:03
1.
/// <summary>
/// 获取查询字符串
/// </summary>
public string GetQueryStr()
{
string str = " From T_PRC_Process as a "
+ " inner join T_BAS_WorkCenter as b on a.WorkCenterID=b.WorkCenterID "
+ " inner join T_PRC_PartSpecPara as c on a.PartSpecParaID=c.PartSpecParaID "
+ " inner join T_BAS_Bearing as d on a.BearingID=d.BearingID "
+ " inner join T_BAS_ForgePrcType as e ON a.FgPrcTypeID = e.FgPrcTypeID "
+ " left join T_PRC_ForgeFactoryProcess as f on a.ProcessID=IsNull(f.ProcessID,'') "
+ " left join T_BAS_ForgeFactory as g on IsNull(f.ForgeFactoryID,'')=IsNull(g.ForgeFactoryID,'') ";
return str;
}
---------------------------------------------------------------------------------------
2.
/// <summary>
/// 计算2条直线的相交点
/// </summary>
public double[] GetLineInt(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4)
{
double[] p1 = new double[3] ;

double x5;
double y5;

if(((x3 - x4) * (y1 - y2) - (x1 - x2) * (y3 - y4)) != 0)
{
x5 = ((x1 - x2) * (x3 * y4 - x4 * y3) - (x3 - x4) * (x1 * y2 - x2 * y1)) / ((x3 - x4) * (y1 - y2) - (x1 - x2) * (y3 - y4));
y5 = ((y1 - y2) * (x3 * y4 - x4 * y3) - (x1 * y2 - x2 * y1) * (y3 - y4)) / ((y1 - y2) * (x3 - x4) - (x1 - x2) * (y3 - y4));

p1[0] = x5;
p1[1] = y5;
p1[2] = 0;
return p1;
}
else
{
return null;
}

}
--------------------------------------------------------
3.
/// <summary>
/// 描 述:获取线的角度
/// </summary>
public double GetLineAngle(double[] lineStartPoint,double[] lineEndPoint)
{
if (lineStartPoint[0]==lineEndPoint[0])
{
return Math.PI/2;
}
else
{
return Math.Atan( (lineStartPoint[1]-lineEndPoint[1])/(lineStartPoint[0]-lineEndPoint[0]) );
}
}
----------------------------------------------------------------
4.
/// <summary>
/// 描 述:获取一个点在线上的垂直点位置
/// </summary>
public double[] GetPointToLineVerticalPoint(double[] point,double[] lineStartPoint,double[] lineEndPoint)
{
double[] vp = new double[3];
string lineDirection=GetLineDirection(lineStartPoint,lineEndPoint);//该方法在下边
if (lineDirection=="H")
{//纵线
vp[0]=lineStartPoint[0];
vp[1]=point[1];
}
else if (lineDirection=="W")
{//横线
vp[0]=point[0];
vp[1]=lineStartPoint[1];
}
else
{//斜线
//先求ab的斜率
double k0 =(lineEndPoint[1]-lineStartPoint[1])/(lineEndPoint[0]-lineStartPoint[0]);
//其垂线的斜率为
double k1 = - 1/k0;

//垂足坐标：
double x=(point[1]-lineStartPoint[1]+lineStartPoint[0]*k0-point[0]*k1)/(k0-k1);

double y=point[1]+(x-point[0])*k1;

vp[0]=x;
vp[1]=y;
}
return vp;
}
-----------------------------------------
5.
/// <summary>
/// 描 述:获取线的方向，H 纵线，W 横线，S 斜线
/// </summary>
public string GetLineDirection(double[] p1,double[] p2)
{
if (Math.Round(p1[0],8)==Math.Round(p2[0],8))
{//纵线
return "H";
}
else if (Math.Round(p1[1],8)==Math.Round(p2[1],8))
{//横线
return "W";
}
else
{//斜线
return "S";
}
}
C语言实在是想不起来了，希望XDJM们帮忙
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bobo364 2010-05-24
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像f.ForgeFactoryID这样的不好办是结构体还好办，是方法还要重写函数

/// <summary>

/// 获取查询字符串

/// </summary>

string GetQueryStr()

{

string str = " From T_PRC_Process as a "

+ " inner join T_BAS_WorkCenter as b on a.WorkCenterID=b.WorkCenterID "  

+ " inner join T_PRC_PartSpecPara as c on a.PartSpecParaID=c.PartSpecParaID "

+ " inner join T_BAS_Bearing as d on a.BearingID=d.BearingID "

+ " inner join T_BAS_ForgePrcType as e ON a.FgPrcTypeID = e.FgPrcTypeID "

+ " left join T_PRC_ForgeFactoryProcess as f on a.ProcessID=IsNull(f.ProcessID,'') "

+ " left join T_BAS_ForgeFactory as g on IsNull(f.ForgeFactoryID,'')=IsNull(g.ForgeFactoryID,'') ";

return str;

}
huanmie_09 2010-05-24
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#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<math.h>



#define PI 3.1415926535897931



char *GetQueryStr()

{

	char *str;

	str = (char *)malloc(1024*sizeof(char));

	if(str == NULL) {

		printf("malloc error!\n");

		return NULL;

	}

	sprintf(str, " From T_PRC_Process as a "

		" inner join T_BAS_WorkCenter as b on a.WorkCenterID=b.WorkCenterID " 

		" inner join T_PRC_PartSpecPara as c on a.PartSpecParaID=c.PartSpecParaID "

		" inner join T_BAS_Bearing as d on a.BearingID=d.BearingID "

		" inner join T_BAS_ForgePrcType as e ON a.FgPrcTypeID = e.FgPrcTypeID "

		" left join T_PRC_ForgeFactoryProcess as f on a.ProcessID=IsNull(f.ProcessID,'') "

		" left join T_BAS_ForgeFactory as g on IsNull(f.ForgeFactoryID,'')=IsNull(g.ForgeFactoryID,'') ");

	return str;

}



/// <summary>

/// 计算2条直线的相交点

/// </summary>

double *GetLineInt(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4)

{

	double *p1 = (double *)malloc(3*sizeof(double)); 

	if(p1 == NULL) {

		printf("malloc error!\n");

		return NULL;

	}

	double x5, y5;

	if(((x3 - x4) * (y1 - y2) - (x1 - x2) * (y3 - y4)) != 0) {

		x5 = ((x1 - x2) * (x3 * y4 - x4 * y3) - (x3 - x4) * (x1 * y2 - x2 * y1)) / ((x3 - x4) * (y1 - y2) - (x1 - x2) * (y3 - y4));

		y5 = ((y1 - y2) * (x3 * y4 - x4 * y3) - (x1 * y2 - x2 * y1) * (y3 - y4)) / ((y1 - y2) * (x3 - x4) - (x1 - x2) * (y3 - y4));



		p1[0] = x5;

		p1[1] = y5;

		p1[2] = 0;

		return p1;

	}

	else {

		free(p1);

		return NULL;

	}

}



/// <summary>

/// 描    述:获取线的角度

/// </summary>

double GetLineAngle(double *lineStartPoint,double *lineEndPoint)

{

	if(lineStartPoint[0]==lineEndPoint[0]) {

		return PI/2;

	}

	else {

		return atan( (lineStartPoint[1]-lineEndPoint[1])/(lineStartPoint[0]-lineEndPoint[0]) );

	}

}



/// <summary>

/// 描    述:获取线的方向，H 纵线，W 横线，S 斜线

/// </summary>

char GetLineDirection(double *p1,double *p2)

{



 

	if (fabs(p1[0] - p2[0]) < 1e-9 ) {//纵线

		return 'H';

	}

	else if (fabs(p1[1] - p2[2]) < 1e-9 ) {//横线

		return 'W';

	}

	else {//斜线

		return 'S';

	}

}



/// <summary>

/// 描    述:获取一个点在线上的垂直点位置

/// </summary>

double *GetPointToLineVerticalPoint(double *point,double *lineStartPoint,double *lineEndPoint)

{

	double *vp = (double *)malloc(3*sizeof(double)); 

	char lineDirection;

	if(vp == NULL) {

		printf("malloc error!\n");

		return NULL;

	}

	lineDirection = GetLineDirection(lineStartPoint,lineEndPoint);

    if (lineDirection=='H') {//纵线

		vp[0]=lineStartPoint[0];

		vp[1]=point[1];

	}

	else if (lineDirection=='W') {//横线

		vp[0]=point[0];

		vp[1]=lineStartPoint[1];

	}

	else {//斜线

		//先求ab的斜率 

		double k0 =(lineEndPoint[1]-lineStartPoint[1])/(lineEndPoint[0]-lineStartPoint[0]); 

		//其垂线的斜率为 

		double k1 = - 1/k0; 



		//垂足坐标： 

		double x=(point[1]-lineStartPoint[1]+lineStartPoint[0]*k0-point[0]*k1)/(k0-k1); 



		double y=point[1]+(x-point[0])*k1;



		vp[0]=x;

		vp[1]=y;

	}

	return vp;

}
zhangyafei13 2010-05-24
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看看，学习学习
摩尔信使MThings 2010-05-24
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。。。。。。。。。。。。。。。。。。。。。。。。。
dopsop110 2010-05-24
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后面几个可能大家对C#的这种计算方法不太懂，给出注释


     Math.PI  //圆的派值，就是3.145555那个

     Math.Atan  //返回正切值为指定数字的角度

     Math.Round  //舍弃小数的方法，第一个是小数值，第二个是精确到多少位

希望大家帮帮忙啊 谢谢啦
wzywsk 2010-05-24
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可以反汇编后在转？
yangyunzhao 2010-05-24
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楼上的网友，你们不是在帮楼主，而是在还楼主。
就是因为有了这么多“热心”网友，很多人上课就不认真了，“反正有人帮忙弄嘛”，作业不自己完成，能学到东西么？
yanran_hill 2010-05-24
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1.

//不考虑线程安全

char * GetQueryStr()

{

    static char str[] = " From T_PRC_Process as a "

                 " inner join T_BAS_WorkCenter as b on"

                 " a.WorkCenterID=b.WorkCenterID " 

                " inner join T_PRC_PartSpecPara as c on "

                " a.PartSpecParaID=c.PartSpecParaID "

                " inner join T_BAS_Bearing as d on a.BearingID=d.BearingID "

                " inner join T_BAS_ForgePrcType as e ON a.FgPrcTypeID ="

                " e.FgPrcTypeID "

                " left join T_PRC_ForgeFactoryProcess as f on a.ProcessID="

                " IsNull(f.ProcessID,'') "

                " left join T_BAS_ForgeFactory as g on ""

                "IsNull(f.ForgeFactoryID,'')=IsNull(g.ForgeFactoryID,'') ";

    return str;

}

-------------------------------------------------------------------------------------

2.

double* GetLineInt(double x1,double y1,double x2,double y2,double x3,

                   double y3,double x4,double y4)

{

    double * p1; 



    double x5;

    double y5;



    p1 = (double *)malloc(sizeof(double)*3);



    if(((x3 - x4) * (y1 - y2) - (x1 - x2) * (y3 - y4)) != 0)

    {

        x5 = ((x1 - x2) * (x3 * y4 - x4 * y3) - (x3 - x4) * (x1 * y2 - x2 * y1)) /

             ((x3 - x4) * (y1 - y2) - (x1 - x2) * (y3 - y4));

        y5 = ((y1 - y2) * (x3 * y4 - x4 * y3) - (x1 * y2 - x2 * y1) * (y3 - y4)) /

            ((y1 - y2) * (x3 - x4) - (x1 - x2) * (y3 - y4));



        p1[0] = x5;

        p1[1] = y5;

        p1[2] = 0;

        return p1;

    }

    else

    {

        delete[] p1;//需要释放空间

         return NULL;

    }

}
dopsop110 2010-05-24
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哦 继续顶啊。。。。分不够可以加
thanktheworld 2010-05-24
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小弟，你真厉害，佩服，我让同事把分都给你了。谢谢哈！
dopsop110 2010-05-24
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谢谢你啊
huanmie_09 2010-05-24
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大姐，
打印浮点数,用格式%f好不好.
for(int i=0;i<3;i++){
printf("%s",p1[i]); -->printf("%f",p1[i]);
}
另外，调用完double *p1=GetLineInt(15.0,11.0,20.0,12.0,15.0,20.0,30.0,13.0);
之后，最好判断一段p1是否为NULL,因为函数可能返回为NULL,如果为NULL的话，后面的循环打印就会出错了.
thanktheworld 2010-05-24
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huanmie_09你不要分了吗，第二个问题你解决了，分就全给你了，要不别人给你解决了，都给别人了不太可惜了么。
thanktheworld 2010-05-24
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#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

#define PI 3.1415926535897931

double *GetLineInt(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4)
{
double *p1 = (double *)malloc(3*sizeof(double));
if(p1 == NULL) {
printf("malloc error!\n");
return NULL;
}

double x5, y5;
if(((x3 - x4) * (y1 - y2) - (x1 - x2) * (y3 - y4)) != 0) {
x5 = ((x1 - x2) * (x3 * y4 - x4 * y3) - (x3 - x4) * (x1 * y2 - x2 * y1)) / ((x3 - x4) * (y1 - y2) - (x1 - x2) * (y3 - y4));
y5 = ((y1 - y2) * (x3 * y4 - x4 * y3) - (x1 * y2 - x2 * y1) * (y3 - y4)) / ((y1 - y2) * (x3 - x4) - (x1 - x2) * (y3 - y4));

p1[0] = x5;
p1[1] = y5;
p1[2] = 0;
return p1;
}
else {
free(p1);
return NULL;
}
}

int main(int argc, char *argv[])
{
double *p1=GetLineInt(15.0,11.0,20.0,12.0,15.0,20.0,30.0,13.0);
for(int i=0;i<3;i++){
printf("%s",p1[i]);
}
system("pause");
}

huanmie_09你试试

thanktheworld 2010-05-24
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huanmie_09，你的第二个方法好像有问题啊，我分别在.net和c环境里测试了一下，得出的结果不一样啊。你看看是哪里出毛病了。
renbin5566 2010-05-23
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lu guo
cattycat 2010-05-23
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后面的需要math的库的，可能需要#include <cmath>
cattycat 2010-05-23
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第一个去掉public就可以了。第二个返回数组的改成指针。其他的类似。

 string GetQueryStr()

        {

            string str = " From T_PRC_Process as a "

            + " inner join T_BAS_WorkCenter as b on a.WorkCenterID=b.WorkCenterID " 

            + " inner join T_PRC_PartSpecPara as c on a.PartSpecParaID=c.PartSpecParaID "

            + " inner join T_BAS_Bearing as d on a.BearingID=d.BearingID "

            + " inner join T_BAS_ForgePrcType as e ON a.FgPrcTypeID = e.FgPrcTypeID "

            + " left join T_PRC_ForgeFactoryProcess as f on a.ProcessID=IsNull(f.ProcessID,'') "

            + " left join T_BAS_ForgeFactory as g on IsNull(f.ForgeFactoryID,'')=IsNull(g.ForgeFactoryID,'') ";

            return str;

        }

---------------------------------------------------------------------------------------

2.

        /// <summary>

        /// 计算2条直线的相交点

        /// </summary>

        double* GetLineInt(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4)

        {

            double* p1 = new double[3] ; 



            double x5;

            double y5;



            if(((x3 - x4) * (y1 - y2) - (x1 - x2) * (y3 - y4)) != 0)

            {

                x5 = ((x1 - x2) * (x3 * y4 - x4 * y3) - (x3 - x4) * (x1 * y2 - x2 * y1)) / ((x3 - x4) * (y1 - y2) - (x1 - x2) * (y3 - y4));

                y5 = ((y1 - y2) * (x3 * y4 - x4 * y3) - (x1 * y2 - x2 * y1) * (y3 - y4)) / ((y1 - y2) * (x3 - x4) - (x1 - x2) * (y3 - y4));



                p1[0] = x5;

                p1[1] = y5;

                p1[2] = 0;

                return p1;

            }

            else

            {

                delete[] p1;//需要释放空间

                return NULL;

            }



        }
zhaiershuaixu 2010-05-23
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不为jf，帮你顶了，下学期可能开c#.
qifeifei 2010-05-23
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楼主啊，这个任务大了啊，你给我多少分啊。
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