菜鸟求解。。角度转弧度

[334628177]

一个角度转弧度的方法要如何构建 我的老是有错误

static double ddmmssToRad(double dms);

int dd=(int)dms;
int mm = (int)(dms - dd) * 100;
double ss = (dms -dd-mm ) * 10000;
double dms = (double)(dd + mm / 60 + ss / 3600)*pi/180;



return double dms;

[some_nothing]

int mm = (int)(dms - dd) * 100;
改成int mm = (int)((dms - dd) * 100)试试,具体未测试

[some_nothing]

不好意思~ 是你的格式错了好多
static double ddmmssToRad(double dms)
{
int dd = (int)dms;
int mm = (int)(dms - dd) * 100;
double ss = (dms - dd - mm) * 10000;
dms = (double)(dd + mm / 60 + ss / 3600) * 3.1415926 / 180;
return dms;
}