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>();
}
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)
When using long double
s 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))
The output is:
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)
Note that this is only using double
s.
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?