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

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