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std::frexp,std::frexpf,std::frexpl(3) C++ Standard Libary std::frexp,std::frexpf,std::frexpl(3)

NAME

std::frexp,std::frexpf,std::frexpl - std::frexp,std::frexpf,std::frexpl

Synopsis


Defined in header <cmath>
float frexp ( float num, int* exp );


double frexp ( double num, int* exp ); (until C++23)


long double frexp ( long double num, int* exp );
constexpr /* floating-point-type */
frexp ( /* floating-point-type */ num, (since C++23)
int* exp ); (1)
float frexpf( float num, int* exp ); (2) (since C++11)
(constexpr since C++23)
long double frexpl( long double num, int* exp ); (3) (since C++11)
(constexpr since C++23)
Additional overloads (since C++11)
Defined in header <cmath>
template< class Integer > (A) (constexpr since C++23)
double frexp ( Integer num, int* exp );


1-3) Decomposes given floating point value num into a normalized fraction and an
integral exponent of two.
The library provides overloads of std::frexp for all cv-unqualified floating-point
types as the type of the parameter num.
(since C++23)


A) Additional overloads are provided for all integer types, which are (since C++11)
treated as double.

Parameters


num - floating-point or integer value
exp - pointer to integer value to store the exponent to

Return value


If num is zero, returns zero and stores zero in *exp.


Otherwise (if num is not zero), if no errors occur, returns the value x in the range
(-1, -0.5], [0.5, 1) and stores an integer value in *exp such that x×2(*exp)
== num.


If the value to be stored in *exp is outside the range of int, the behavior is
unspecified.

Error handling


This function is not subject to any errors specified in math_errhandling.


If the implementation supports IEEE floating-point arithmetic (IEC 60559),


* If num is ±0, it is returned, unmodified, and 0 is stored in *exp.
* If num is ±∞, it is returned, and an unspecified value is stored in *exp.
* If num is NaN, NaN is returned, and an unspecified value is stored in *exp.
* No floating-point exceptions are raised.
* If FLT_RADIX is 2 (or a power of 2), the returned value is exact, the current
rounding mode is ignored.

Notes


On a binary system (where FLT_RADIX is 2), std::frexp may be implemented as


{
*exp = (value == 0) ? 0 : (int)(1 + std::logb(value));
return std::scalbn(value, -(*exp));
}


The function std::frexp, together with its dual, std::ldexp, can be used to
manipulate the representation of a floating-point number without direct bit
manipulations.


The additional overloads are not required to be provided exactly as (A). They only
need to be sufficient to ensure that for their argument num of integer type,
std::frexp(num, exp) has the same effect as std::frexp(static_cast<double>(num),
exp).

Example


Compares different floating-point decomposition functions:

// Run this code


#include <cmath>
#include <iostream>
#include <limits>


int main()
{
double f = 123.45;
std::cout << "Given the number " << f << " or " << std::hexfloat
<< f << std::defaultfloat << " in hex,\n";


double f3;
double f2 = std::modf(f, &f3);
std::cout << "modf() makes " << f3 << " + " << f2 << '\n';


int i;
f2 = std::frexp(f, &i);
std::cout << "frexp() makes " << f2 << " * 2^" << i << '\n';


i = std::ilogb(f);
std::cout << "logb()/ilogb() make " << f / std::scalbn(1.0, i)
<< " * " << std::numeric_limits<double>::radix
<< "^" << std::ilogb(f) << '\n';
}

Possible output:


Given the number 123.45 or 0x1.edccccccccccdp+6 in hex,
modf() makes 123 + 0.45
frexp() makes 0.964453 * 2^7
logb()/ilogb() make 1.92891 * 2^6

See also


ldexp
ldexpf multiplies a number by 2 raised to an integral power
ldexpl (function)
(C++11)
(C++11)
logb
logbf
logbl extracts exponent of the number
(C++11) (function)
(C++11)
(C++11)
ilogb
ilogbf
ilogbl extracts exponent of the number
(C++11) (function)
(C++11)
(C++11)
modf
modff decomposes a number into integer and fractional parts
modfl (function)
(C++11)
(C++11)
C documentation for
frexp

2024.06.10 http://cppreference.com