本文整理汇总了C++中BigFloat::sub方法的典型用法代码示例。如果您正苦于以下问题：C++ BigFloat::sub方法的具体用法？C++ BigFloat::sub怎么用？C++ BigFloat::sub使用的例子？那么, 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BigFloat的用法示例。
在下文中一共展示了BigFloat::sub方法的2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: invsqrt
BigFloat invsqrt(uint32_t x,size_t p,int tds){
// Compute inverse square root using Newton's Method.
// ( r0^2 * x - 1 )
// r1 = r0 - (----------------) * r0
// ( 2 )
if (x == 0)
throw "Divide by Zero";
// End of recursion. Generate starting point.
if (p == 0){
double val = 1. / sqrt((double)x);
int64_t exponent = 0;
// Scale
while (val < 1000000000.){
val *= 1000000000.;
exponent--;
}
// Rebuild a BigFloat.
uint64_t val64 = (uint64_t)val;
BigFloat out;
out.sign = true;
out.T = std::unique_ptr<uint32_t[]>(new uint32_t[2]);
out.T[0] = (uint32_t)(val64 % 1000000000);
out.T[1] = (uint32_t)(val64 / 1000000000);
out.L = 2;
out.exp = exponent;
return out;
}
// Half the precision
size_t s = p / 2 + 1;
if (p == 1) s = 0;
if (p == 2) s = 1;
// Recurse at half the precision
BigFloat T = invsqrt(x,s,tds);
BigFloat temp = T.mul(T,p); // r0^2
temp = temp.mul(x,p,tds); // r0^2 * x
temp = temp.sub(BigFloat(1),p); // r0^2 * x - 1
temp = temp.mul(500000000); // (r0^2 * x - 1) / 2
temp.exp--;
temp = temp.mul(T,p,tds); // (r0^2 * x - 1) / 2 * r0
return T.sub(temp,p); // r0 - (r0^2 * x - 1) / 2 * r0
}示例2: rcp
BigFloat BigFloat::rcp(size_t p,int tds) const{
// Compute reciprocal using Newton's Method.
// r1 = r0 - (r0 * x - 1) * r0
if (L == 0)
throw "Divide by Zero";
// Collect operand
int64_t Aexp = exp;
size_t AL = L;
uint32_t *AT = T.get();
// End of recursion. Generate starting point.
if (p == 0){
// Truncate precision to 3.
p = 3;
if (AL > p){
size_t chop = AL - p;
AL = p;
Aexp += chop;
AT += chop;
}
// Convert number to floating-point.
double val = AT[0];
if (AL >= 2)
val += AT[1] * 1000000000.;
if (AL >= 3)
val += AT[2] * 1000000000000000000.;
// Compute reciprocal.
val = 1. / val;
Aexp = -Aexp;
// Scale
while (val < 1000000000.){
val *= 1000000000.;
Aexp--;
}
// Rebuild a BigFloat.
uint64_t val64 = (uint64_t)val;
BigFloat out;
out.sign = sign;
out.T = std::unique_ptr<uint32_t[]>(new uint32_t[2]);
out.T[0] = (uint32_t)(val64 % 1000000000);
out.T[1] = (uint32_t)(val64 / 1000000000);
out.L = 2;
out.exp = Aexp;
return out;
}
// Half the precision
size_t s = p / 2 + 1;
if (p == 1) s = 0;
if (p == 2) s = 1;
// Recurse at half the precision
BigFloat T = rcp(s,tds);
// r1 = r0 - (r0 * x - 1) * r0
return T.sub(this->mul(T,p,tds).sub(BigFloat(1),p).mul(T,p,tds),p);
}