用ldexp反转frexp [英] Inverting frexp with ldexp
问题描述
如果我了解文档
If I understand the documentation correctly, we should be able to use ldexp to recover a floating point number decomposed into a signed mantissa and an exponent by frexp. I have been unable to achieve this. Consider the following code:
#include <cmath>
#include <iostream>
#include <limits>
template <typename T>
void float_info() {
std::cout << "max=" << std::numeric_limits<T>::max() <<
", max_exp=" << std::numeric_limits<T>::max_exponent <<
", max_10_exp=" << std::numeric_limits<T>::max_exponent10 <<
", min=" << std::numeric_limits<T>::min() <<
", min_exp=" << std::numeric_limits<T>::min_exponent <<
", min_10_exp=" << std::numeric_limits<T>::min_exponent10 <<
", dig=" << std::numeric_limits<T>::digits10 <<
", mant_dig=" << std::numeric_limits<T>::digits <<
", epsilon=" << std::numeric_limits<T>::epsilon() <<
", radix=" << std::numeric_limits<T>::radix <<
", rounds=" << std::numeric_limits<T>::round_style << std::endl;
}
template <typename T>
void compare(T a, T b) {
std::cout << a << " " << b << " (" <<
(a != b ? "un" : "") << "equal)" << std::endl;
}
template<typename T>
void test_ldexp() {
float_info<T>();
T x = 1 + std::numeric_limits<T>::epsilon();
T y = ldexp(x, 0);
int exponent;
T mantissa = frexp(x, &exponent);
T z = ldexp(mantissa, exponent);
compare(x, y);
compare(x, z);
std::cout << std::endl;
}
int main() {
std::cout.precision(25);
test_ldexp<float>();
test_ldexp<double>();
test_ldexp<long double>();
}
在Ubuntu 14.04.3 LTS上使用g++(版本4.8.4)进行编译时，输出为:
When compiled with g++ (version 4.8.4) on Ubuntu 14.04.3 LTS, the output is:
max=3.402823466385288598117042e+38, max_exp=128, max_10_exp=38,
min=1.175494350822287507968737e-38, min_exp=-125, min_10_exp=-37, dig=6,
mant_dig=24, epsilon=1.1920928955078125e-07, radix=2, rounds=1
1.00000011920928955078125 1.00000011920928955078125 (equal)
1.00000011920928955078125 1.00000011920928955078125 (equal)
max=1.797693134862315708145274e+308, max_exp=1024, max_10_exp=308,
min=2.225073858507201383090233e-308, min_exp=-1021, min_10_exp=-307, dig=15,
mant_dig=53, epsilon=2.220446049250313080847263e-16, radix=2, rounds=1
1.000000000000000222044605 1.000000000000000222044605 (equal)
1.000000000000000222044605 1.000000000000000222044605 (equal)
max=1.189731495357231765021264e+4932, max_exp=16384, max_10_exp=4932,
min=3.362103143112093506262678e-4932, min_exp=-16381, min_10_exp=-4931, dig=18,
mant_dig=64, epsilon=1.084202172485504434007453e-19, radix=2, rounds=1
1.00000000000000000010842 1 (unequal)
1.00000000000000000010842 1 (unequal)
使用long double时，我们似乎通过将x与frexpr分解而丢失了一些东西.如果使用python3(版本3.4.3)运行以下脚本，则可以实现我期望的行为.
When using long doubles we appear to be losing something by decomposing our x with frexpr. I can achieve the behavior I expect if I run the following script with python3 (version 3.4.3).
import math
import sys
def compare(a, b):
print('{a} {b} ({pre}equal)'.format(a=a, b=b,
pre='un' if a != b else ''))
x = 1 + sys.float_info.epsilon
mantissa, exponent = math.frexp(x)
print(sys.float_info)
compare(x, math.ldexp(x, 0))
compare(x, math.ldexp(mantissa, exponent))
输出为:
sys.float_info(max=1.7976931348623157e+308, max_exp=1024, max_10_exp=308,
min=2.2250738585072014e-308, min_exp=-1021, min_10_exp=-307, dig=15,
mant_dig=53, epsilon=2.220446049250313e-16, radix=2, rounds=1)
1.0000000000000002 1.0000000000000002 (equal)
1.0000000000000002 1.0000000000000002 (equal)
请注意，这仅使用double.
我试图读取cmath头文件，以了解frexpr和ldexpr的实现方式，但我无法理解.发生了什么事?
I tried to read the cmath header file to understand how frexpr and ldexpr are implemented, but I couldn't make sense of it. What is going on?
推荐答案
通过包括typeinfo标头并在test_ldexp函数中对compare的调用之前插入以下行，来修改C ++程序.
Modify your C++ program by including the typeinfo header and inserting the following lines right before the calls to compare in your test_ldexp function.
std::cout << "types:" << std::endl;
std::cout << " x : " << typeid(x).name() << std::endl;
std::cout << " mantissa : " << typeid(frexp(x, &exponent)).name()
<< std::endl;
std::cout << " ldexp(...): " << typeid(ldexp(x, 0)).name() << std::endl;
std::cout << " ldexp(...): " << typeid(ldexp(mantissa, exponent)).name()
<< std::endl;
现在输出为:
max=3.402823466385288598117042e+38, max_exp=128, max_10_exp=38,
min=1.175494350822287507968737e-38, min_exp=-125, min_10_exp=-37, dig=6,
mant_dig=24, epsilon=1.1920928955078125e-07, radix=2, rounds=1
types:
x : f
mantissa : d
ldexp(...): d
ldexp(...): d
1.00000011920928955078125 1.00000011920928955078125 (equal)
1.00000011920928955078125 1.00000011920928955078125 (equal)
max=1.797693134862315708145274e+308, max_exp=1024, max_10_exp=308,
min=2.225073858507201383090233e-308, min_exp=-1021, min_10_exp=-307, dig=15,
mant_dig=53, epsilon=2.220446049250313080847263e-16, radix=2, rounds=1
types:
x : d
mantissa : d
ldexp(...): d
ldexp(...): d
1.000000000000000222044605 1.000000000000000222044605 (equal)
1.000000000000000222044605 1.000000000000000222044605 (equal)
max=1.189731495357231765021264e+4932, max_exp=16384, max_10_exp=4932,
min=3.362103143112093506262678e-4932, min_exp=-16381, min_10_exp=-4931, dig=18,
mant_dig=64, epsilon=1.084202172485504434007453e-19, radix=2, rounds=1
types:
x : e
mantissa : d
ldexp(...): d
ldexp(...): d
1.00000000000000000010842 1 (unequal)
1.00000000000000000010842 1 (unequal)
无论输入什么类型，
frexpr和ldexpr都将返回double！看来您正在使用math.h中定义的功能(请参见此处和此处)，而不是在cmath中定义的内容.将对frexpr和ldexpr的调用替换为对std::frexpr和std::ldexpr的调用，您的代码将按预期工作.
frexpr and ldexpr are returning doubles no matter what type you put in! It appears that you are using the functions defined in math.h (see here and here) instead those defined in cmath. Replace your calls to frexpr and ldexpr with calls to std::frexpr and std::ldexpr and your code will work as you expect.
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