subject_18341.htm
来自「一些关于vc的问答」· HTM 代码 · 共 120 行 · 第 1/2 页
HTM
120 行
<p>
序号：18341 发表者：西门吹雪 发表日期：2002-10-19 21:19:23
<br>主题：大家来运行这个程序试试,结果不是我所要的,问题在哪里
<br>内容：很简单的一个小程序:<BR>#include "iostream.h"<BR>#include "math.h"<BR>#define PI 3.14159265358979323846<BR>#define RADIANS(x) PI*(x)/180.0<BR>#define ARADIANS(x) 180.0*(x)/PI<BR>#define D1 7.83<BR>#define D2 7.03<BR>#define D3 6.51<BR>#define D4 0.0<BR>#define D5 0.0<BR>#define D6 2.63<BR>typedef struct adver<BR>{<BR> double tht1,tht2,tht3,tht4,tht5,tht6;<BR>}ADVERSE;<BR><BR>ADVERSE AdverEvaluate(double px,double py,double pz,double roll,double pitch,double yaw)<BR>{<BR> ADVERSE adver;<BR> double tht1(0.0),tht2(0.0),tht3(0.0),tht4(0.0),tht5(0.0),tht6(0.0);<BR> double p3x,p3y,p3z,p1,p2;<BR> double ax,ay,az,nx,ny,nz,ox,oy,oz;<BR> double T11,T12,T13,T14,T21,T22,T23,T24,T31,T32,T33,T34, T;<BR> double a11,a12,a13,a14,a21,a22,a23,a24,a31,a32,a33,a34;<BR><BR> nx=cos(roll)*cos(pitch);<BR> ny=sin(roll)*cos(pitch);<BR> nz=-sin(pitch);<BR><BR> ox=cos(roll)*sin(pitch)*sin(yaw)-sin(roll)*cos(yaw);<BR> oy=sin(roll)*sin(pitch)*sin(yaw)+cos(roll)*cos(yaw);<BR> oz=cos(pitch)*sin(yaw);<BR><BR> ax=cos(roll)*sin(pitch)*cos(yaw)+sin(roll)*sin(yaw);<BR> ay=sin(roll)*sin(pitch)*cos(yaw)-cos(roll)*sin(yaw);<BR> az=cos(pitch)*cos(yaw);<BR><BR> p3x=px-D6*ax;<BR> p3y=py-D6*ay;<BR> p3z=py-D6*az;<BR><BR> a11=nx*cos(tht1)*cos(tht2)+ny*sin(tht1)*cos(tht2)+nz*sin(tht2);<BR> a12=ox*cos(tht1)*cos(tht2)+oy*sin(tht1)*cos(tht2)+oz*sin(tht2);<BR> a13=ax*cos(tht1)*cos(tht2)+ay*sin(tht1)*cos(tht2)+az*sin(tht2);<BR> a14=p3x*cos(tht1)*cos(tht2)+p3y*sin(tht1)*cos(tht2)+p3z-D1*sin(tht2)-D2;<BR> <BR> a21=-nx*cos(tht1)*sin(tht2)-ny*sin(tht1)*cos(tht2)+nz*cos(tht2);<BR> a22=-ox*cos(tht1)*sin(tht2)-oy*sin(tht1)*cos(tht2)+oz*cos(tht2);<BR> a23=-ax*cos(tht1)*sin(tht2)-ay*sin(tht1)*cos(tht2)+az*cos(tht2);<BR> a24=-p3x*cos(tht1)*sin(tht2)-p3y*sin(tht1)*cos(tht2)+p3z*cos(tht2)-D1*cos(tht2);<BR><BR> a31=nx*sin(tht1)-ny*cos(tht1);<BR> a32=ox*sin(tht1)-oy*cos(tht1);<BR> a33=ax*sin(tht1)-ay*cos(tht1);<BR> a34=p3x*sin(tht1)-p3y*cos(tht1);<BR><BR> tht1=atan2(p3x,p3y);<BR><BR> p1=p3x*cos(tht1)-p3y*sin(tht1);<BR> p2=p3z-D1;<BR> T=(p1*p1+p2*p2-D2*D2-D3*D3)/(2*D2*D3);//T=cos(tht3);<BR><BR> tht3=atan2(sqrt(1-T*T),T);<BR> tht2=atan2((p2*(D3*cos(tht3)+D2)-p1*D3*sin(tht3)),(p1*(D3*cos(tht3)+D2)-p2*D3*sin(tht3)));<BR><BR> T11=nx*(sin(tht3)*cos(tht1)*cos(tht2)*sin(tht3)*cos(tht1)*cos(tht2)+cos(tht1)*cos(tht3)*sin(tht2)*cos(tht1)*cos(tht3)*sin(tht2))+<BR> ny*(sin(tht1)*sin(tht3)*cos(tht2)*sin(tht1)*sin(tht3)*cos(tht2)+sin(tht1)*sin(tht2)*cos(tht3)*sin(tht1)*sin(tht2)*cos(tht3))+<BR> nz*(sin(tht3)*sin(tht2)*sin(tht3)*sin(tht2)+cos(tht3)*cos(tht3)*cos(tht2)*cos(tht2));<BR> T12=ox*(sin(tht3)*cos(tht1)*cos(tht2)*sin(tht3)*cos(tht1)*cos(tht2)+cos(tht1)*cos(tht3)*sin(tht2)*cos(tht1)*cos(tht3)*sin(tht2))+<BR> oy*(sin(tht1)*sin(tht3)*cos(tht2)*sin(tht1)*sin(tht3)*cos(tht2)+sin(tht1)*sin(tht2)*cos(tht3)*sin(tht1)*sin(tht2)*cos(tht3))+<BR> oz*(sin(tht3)*sin(tht2)*sin(tht3)*sin(tht2)+cos(tht3)*cos(tht3)*cos(tht2)*cos(tht2));<BR> T13=ax*(sin(tht3)*cos(tht1)*cos(tht2)*sin(tht3)*cos(tht1)*cos(tht2)+cos(tht1)*cos(tht3)*sin(tht2)*cos(tht1)*cos(tht3)*sin(tht2))+<BR> ay*(sin(tht1)*sin(tht3)*cos(tht2)*sin(tht1)*sin(tht3)*cos(tht2)+sin(tht1)*sin(tht2)*cos(tht3)*sin(tht1)*sin(tht2)*cos(tht3))+<BR> az*(sin(tht3)*sin(tht2)*sin(tht3)*sin(tht2)+cos(tht3)*cos(tht3)*cos(tht2)*cos(tht2));<BR> T14=px*(sin(tht3)*cos(tht1)*cos(tht2)*sin(tht3)*cos(tht1)*cos(tht2)+cos(tht1)*cos(tht3)*sin(tht2)*cos(tht1)*cos(tht3)*sin(tht2))+<BR> py*(sin(tht1)*sin(tht3)*cos(tht2)*sin(tht1)*sin(tht3)*cos(tht2)+sin(tht1)*sin(tht2)*cos(tht3)*sin(tht1)*sin(tht2)*cos(tht3))+<BR> pz*(sin(tht3)*sin(tht2)*sin(tht3)*sin(tht2)+cos(tht3)*cos(tht3)*cos(tht2)*cos(tht2));<BR> T21=nx*sin(tht1)*sin(tht1)+ny*cos(tht1)*cos(tht1);<BR> T22=ox*sin(tht1)*sin(tht1)+oy*cos(tht1)*cos(tht1);<BR> T23=ax*sin(tht1)*sin(tht1)+ay*cos(tht1)*cos(tht1);<BR> T24=px*sin(tht1)*sin(tht1)+py*cos(tht1)*cos(tht1);<BR> T31=nx*(cos(tht1)*cos(tht2)*cos(tht3)*cos(tht1)*cos(tht2)*cos(tht3)+cos(tht1)*sin(tht2)*sin(tht3)*cos(tht1)*sin(tht2)*sin(tht3))+<BR> ny*(cos(tht2)*sin(tht1)*cos(tht3)*cos(tht2)*sin(tht1)*cos(tht3)+sin(tht1)*sin(tht2)*sin(tht3)*sin(tht1)*sin(tht2)*sin(tht3))+<BR> nz*(cos(tht3)*sin(tht2)*cos(tht3)*sin(tht2)+sin(tht3)*cos(tht2)*sin(tht3)*cos(tht2));<BR> T32=ox*(cos(tht1)*cos(tht2)*cos(tht3)*cos(tht1)*cos(tht2)*cos(tht3)+cos(tht1)*sin(tht2)*sin(tht3)*cos(tht1)*sin(tht2)*sin(tht3))+<BR> oy*(cos(tht2)*sin(tht1)*cos(tht3)*cos(tht2)*sin(tht1)*cos(tht3)+sin(tht1)*sin(tht2)*sin(tht3)*sin(tht1)*sin(tht2)*sin(tht3))+<BR> oz*(cos(tht3)*sin(tht2)*cos(tht3)*sin(tht2)+sin(tht3)*cos(tht2)*sin(tht3)*cos(tht2));<BR> T33=ax*(cos(tht1)*cos(tht2)*cos(tht3)*cos(tht1)*cos(tht2)*cos(tht3)+cos(tht1)*sin(tht2)*sin(tht3)*cos(tht1)*sin(tht2)*sin(tht3))+<BR> ay*(cos(tht2)*sin(tht1)*cos(tht3)*cos(tht2)*sin(tht1)*cos(tht3)+sin(tht1)*sin(tht2)*sin(tht3)*sin(tht1)*sin(tht2)*sin(tht3))+<BR> az*(cos(tht3)*sin(tht2)*cos(tht3)*sin(tht2)+sin(tht3)*cos(tht2)*sin(tht3)*cos(tht2));<BR> T34=px*(cos(tht1)*cos(tht2)*cos(tht3)*cos(tht1)*cos(tht2)*cos(tht3)+cos(tht1)*sin(tht2)*sin(tht3)*cos(tht1)*sin(tht2)*sin(tht3))+<BR> py*(cos(tht2)*sin(tht1)*cos(tht3)*cos(tht2)*sin(tht1)*cos(tht3)+sin(tht1)*sin(tht2)*sin(tht3)*sin(tht1)*sin(tht2)*sin(tht3))+<BR> pz*(cos(tht3)*sin(tht2)*cos(tht3)*sin(tht2)+sin(tht3)*cos(tht2)*sin(tht3)*cos(tht2));<BR> <BR> tht4=atan2(T23,T13);<BR> tht5=atan2(T31,(cos(tht4)*T11+T21));<BR> tht6=atan2((-sin(tht4)*T11+cos(tht4)*T21),(-sin(tht4)*T12+cos(tht4)*T22));<BR> adver.tht1=ARADIANS(tht1);<BR> adver.tht2=ARADIANS(tht2);<BR> adver.tht3=ARADIANS(tht3);<BR> adver.tht4=ARADIANS(tht4);<BR> adver.tht5=ARADIANS(tht5);<BR> adver.tht6=ARADIANS(tht6);<BR> return adver;<BR>}<BR><BR>void main()<BR>{<BR> double px,py,pz,roll,pitch,yaw;<BR> ADVERSE adver;<BR><BR> cout<<"plaese insert:";<BR> cin>>px>>py>>pz>>roll>>pitch>>yaw;<BR> cout<<":";<BR> cout<<adver.tht1<<" "<<adver.tht2<<" "<<adver.tht3<<" "<<adver.tht4<<" "<<adver.tht5<<" "<<adver.tht6;<BR>}<BR>
<br><a href="javascript:history.go(-1)">返回上页</a><br><a href=http://www.copathway.com/cndevforum/>访问论坛</a></p>
<hr size=1>
<blockquote><p>
回复者：QQ糖 回复日期：2002-10-19 21:30:54
<br>内容：我晕了!
<br>
<a href="javascript:history.go(-1)">返回上页</a><br><a href=http://www.copathway.com/cndevforum/>访问论坛</a></p></blockquote>
<hr size=1>
<blockquote><p>
回复者：西门吹雪 回复日期：2002-10-19 21:38:50
<br>内容：你试试嘛,很简单的.考过去运行一下萝,没有错误的,运行的结果不是我想要的,而且不管什么输入,都只有一个输出.麻烦QQ了.分不够再给了.比较急.
<br>
<a href="javascript:history.go(-1)">返回上页</a><br><a href=http://www.copathway.com/cndevforum/>访问论坛</a></p></blockquote>
<hr size=1>
<blockquote><p>
回复者：塞翁 回复日期：2002-10-20 03:13:13
<br>内容：好象这个 "return adver; " 有点问题;
<br>
<a href="javascript:history.go(-1)">返回上页</a><br><a href=http://www.copathway.com/cndevforum/>访问论坛</a></p></blockquote>
<hr size=1>
<blockquote><p>
回复者：西门吹雪 回复日期：2002-10-20 10:31:47
<br>内容：不会的,结构体都要返回这个的
<br>
<a href="javascript:history.go(-1)">返回上页</a><br><a href=http://www.copathway.com/cndevforum/>访问论坛</a></p></blockquote>
<hr size=1>
<blockquote><p>
<font color=red>答案被接受</font><br>回复者：QQ糖 回复日期：2002-10-20 10:32:37
<br>内容：没有调用计算函数啊:<BR><BR>输入数据完了以后,调用一下就可以了!<BR><BR>main函数代码如下:<BR><BR>其它不用变,只要加一句话就行了,有注释的!<BR><BR><BR>void main()<BR>{<BR> double px,py,pz,roll,pitch,yaw;<BR> ADVERSE adver;<BR><BR> cout<<"plaese insert:";<BR> cin>>px>>py>>pz>>roll>>pitch>>yaw;<BR> adver = AdverEvaluate( px, py, pz, roll, pitch, yaw); //加这句话!<BR> cout<<":";<BR> cout<<adver.tht1<<" "<<adver.tht2<<" "<<adver.tht3<<" "<<adver.tht4<<" "<<adver.tht5<<" "<<adver.tht6;<BR>}<BR><BR><BR>不知道对不对?
<br>
<a href="javascript:history.go(-1)">返回上页</a><br><a href=http://www.copathway.com/cndevforum/>访问论坛</a></p></blockquote>
<hr size=1>
<blockquote><p>
回复者：西门吹雪 回复日期：2002-10-20 10:40:23
<br>内容：呵呵，还是绿猪厉害呀，初步检查正确，还有待更进一步的检查。<BR>谢谢绿猪呀。<BR>关于公式的堆砌你能给一些建议吗？
<br>
<a href="javascript:history.go(-1)">返回上页</a><br><a href=http://www.copathway.com/cndevforum/>访问论坛</a></p></blockquote>
<hr size=1>
<blockquote><p>
回复者：QQ糖 回复日期：2002-10-20 10:46:51
<br>内容：自己好好想想嘛!<BR><BR>我一看到这一堆东西,就没想往下看!<BR><BR>呵呵!
<br>
<a href="javascript:history.go(-1)">返回上页</a><br><a href=http://www.copathway.com/cndevforum/>访问论坛</a></p></blockquote>
<hr size=1>
<blockquote><p>
回复者：西门吹雪 回复日期：2002-10-20 10:52:56
<br>内容：是这样的，我的基础比较差呀。所以有点简单的东西不是很明白的。不过经高手一指点就不一样了。谢谢你呀绿猪。<BR>这个我觉得是公式的堆砌，但我没有更好的办法。
<br>
<a href="javascript:history.go(-1)">返回上页</a><br><a href=http://www.copathway.com/cndevforum/>访问论坛</a></p></blockquote>
<hr size=1>
<blockquote><p>
回复者：QQ糖 回复日期：2002-10-20 10:59:19
<br>内容：呵呵,把模样差不多的写个函数吧!
<br>
<a href="javascript:history.go(-1)">返回上页</a><br><a href=http://www.copathway.com/cndevforum/>访问论坛</a></p></blockquote>
<hr size=1>
⌨️ 快捷键说明
复制代码Ctrl + C
搜索代码Ctrl + F
全屏模式F11
增大字号Ctrl + =
减小字号Ctrl + -
显示快捷键?