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

float       frexp ( float num, int* exp );
double      frexp ( double num, int* exp );
long double frexp ( long double num, int* exp );

constexpr /* floating-point-type */
            frexp ( /* floating-point-type */ num, int* exp );

float       frexpf( float num, int* exp );

long double frexpl( long double num, int* exp );

template< class Integer >
double      frexp ( Integer num, int* exp );

Parameters

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

#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';
}

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