helo我实现了一个用于执行定点关节操作的头文件,如下所示,它执行乘法和除法运算。
#ifndef __fixed_point_header_h__
#define __fixed_point_header_h__
#include <boost/operators.hpp>
#include <limits>
namespace fp {
// FP The fixed point(base) type, should be an integer type
// I Integer part bit count
// F fractional part bit count
template<typename FP, unsigned char I, unsigned char F = std::numeric_limits<FP>::digits - I>
class fixed_point: boost::ordered_field_operators<fp::fixed_point<FP, I, F> >
{
//As the fixed_point class needs 2 to the power of P in several parts for floating point conversions, template recursion using template metaprogramming is used to calculate 2 to the power of F at compile time itself.
template<int P,typename T = void>
struct power2
{
static const long long value = 2 * power2<P-1,T>::value;
};
template <typename P>
struct power2<0, P>
{
static const long long value = 1;
};
// Initializing constructor.
//! This constructor takes a value of type FP and initializes the //internal representation of fixed_point<FP, I, F> with it.
fixed_point(FP value,bool): fixed_(value){ } // initializer list
public:
typedef FP base_type; /// fixed point base type of this fixed_point cass.
static const unsigned char integer_bit_count = I; /// integer part bit count.
static const unsigned char fractional_bit_count = F; /// fractional part bit count.
fixed_point(){ } /// Default constructor.
//Conversion by constructors.
//Integer to Fixed point
template<typename T> fixed_point(T value) : fixed_((FP)value << F)
{
BOOST_CONCEPT_ASSERT((boost::Integer<T>));
}
//floating point to fixed point
fixed_point(float value) :fixed_((FP)(value * power2<F>::value))
{ }
fixed_point(double value) : fixed_((FP)(value * power2<F>::value))
{ }
fixed_point(long double value) : fixed_((FP)(value * power2<F>::value))
{ }
/// Copy constructor,explicit definition
fixed_point(
/// The right hand side.
fixed_point<FP, I, F> const& rhs)
//: fixed_(rhs.fixed_)
{
fixed_ = rhs.fixed_;
}
// Copy assignment operator.
fp::fixed_point<FP, I, F> & operator =(fp::fixed_point<FP, I, F> const& rhs)
{
fp::fixed_point<FP, I, F> temp(rhs);
swap(temp);
return *this; //return by reference
}
/// Exchanges the elements of two fixed_point objects.
void swap(
/// The right hand side.
fp::fixed_point<FP, I, F> & rhs)
{
std::swap(fixed_, rhs.fixed_);
}
/// Multiplication.
fp::fixed_point<FP, I, F> & operator *=(
/// Factor for mutliplication.
fp::fixed_point<FP, I, F> const& factor)
{
fixed_ =
( fixed_ * (factor.fixed_ >> F) ) +
( ( fixed_ * (factor.fixed_ & (power2<F>::value-1) ) ) >> F );
return *this; //return A reference to this object.
}
/// Division.
fp::fixed_point<FP, I, F> & operator /=(
/// Divisor for division.
fp::fixed_point<FP, I, F> const& divisor)
{
fixed_ = ( (fixed_)<< F / divisor.fixed_ )+
( ( fixed_ / (divisor.fixed_ & (power2<F>::value-1) ) )<< F );
return *this; //return A reference to this object.
}
private:
/// The value in fixed point format.
FP fixed_;
};
} // namespace fp
#endif // __fixed_point_header__
用于乘法我提升了所得到的不动点的类型,例如(当两个8位数相乘时,产生16位结果)。它工作正常。
但是当我使用以下测试文件进行划分时 TEST.CPP
#include <stdio.h>
#include <fixed_point_header.h>
int main()
{
fp::fixed_point<int, 15> fp1 = 3.0; // 011 0000 0000 0000 0000
fp::fixed_point<int, 15> fp2 = 3.0; // 100 0000 0000 0000 0000
fp::fixed_point<int, 15> fp3;
fp3 = fp1/fp2;
printf("divided value ==%f\n", (float)fp3 );
}
我总是打算浮点异常错误,有些人请检查一下,看看发生了什么。
答案 0 :(得分:0)
当然这可能导致除以零:
fixed_ = ( (fixed_)<< F / divisor.fixed_ )+
( ( fixed_ / (divisor.fixed_ & (power2<F>::value-1) ) )<< F );
return *this; //return A reference to this object.
如果divisor.fixed_ & (power2<F>::value-1)
为零,则除以零。
我也不相信它是正确的,但我会留给你解决这个问题,因为我实际上并没有考虑过你如何做一个“定点”的分歧。