本文整理汇总了C++中expr::m_e1方法的典型用法代码示例。如果您正苦于以下问题：C++ expr::m_e1方法的具体用法？C++ expr::m_e1怎么用？C++ expr::m_e1使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类expr的用法示例。
在下文中一共展示了expr::m_e1方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: solve
inline bool solve(const expr<division,ref2<var<double>>,ref2<var<double>>>& e, const double& r) noexcept
{
// n/m = 3.1415926 => n=22,m=7
long m[2][2] = {
{1,0},
{0,1}
};
double x = r;
//double startx = r;
long maxden = 100;
long ai;
/* loop finding terms until denom gets too big */
while (m[1][0] * ( ai = (long)x ) + m[1][1] <= maxden) {
long t;
t = m[0][0] * ai + m[0][1];
m[0][1] = m[0][0];
m[0][0] = t;
t = m[1][0] * ai + m[1][1];
m[1][1] = m[1][0];
m[1][0] = t;
if(x==(double)ai) break; // AF: division by zero
x = 1/(x - (double) ai);
if(x>(double)0x7FFFFFFF) break; // AF: representation failure
}
///* now remaining x is between 0 and 1/ai */
///* approx as either 0 or 1/m where m is max that will fit in maxden */
///* first try zero */
//printf("%ld/%ld, error = %e\n", m[0][0], m[1][0], startx - ((double) m[0][0] / (double) m[1][0]));
///* now try other possibility */
//ai = (maxden - m[1][1]) / m[1][0];
//m[0][0] = m[0][0] * ai + m[0][1];
//m[1][0] = m[1][0] * ai + m[1][1];
//printf("%ld/%ld, error = %e\n", m[0][0], m[1][0], startx - ((double) m[0][0] / (double) m[1][0]));
e.m_e1(m[0][0]);
e.m_e2(m[1][0]);
return true;
}