std:: modf, std:: modff, std:: modfl

Parameters

Return value

If no errors occur, returns the fractional part of num with the same sign as num . The integral part is put into the value pointed to by iptr .

The sum of the returned value and the value stored in * iptr gives num (allowing for rounding).

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, ±0 is returned, and ±0 is stored in * iptr .
• If num is ±∞, ±0 is returned, and ±∞ is stored in * iptr .
• If num is NaN, NaN is returned, and NaN is stored in * iptr .
• The returned value is exact, the current rounding mode is ignored.

Notes

This function behaves as if implemented as follows:

double modf(double num, double* iptr)
{
#pragma STDC FENV_ACCESS ON
    int save_round = std::fegetround();
    std::fesetround(FE_TOWARDZERO);
    *iptr = std::nearbyint(num);
    std::fesetround(save_round);
    return std::copysign(std::isinf(num) ? 0.0 : num - (*iptr), num);
}

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 :: modf ( num, iptr ) has the same effect as std :: modf ( static_cast < double > ( num ) , iptr ) .

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';
 
    // special values
    f2 = std::modf(-0.0, &f3);
    std::cout << "modf(-0) makes " << f3 << " + " << f2 << '\n';
    f2 = std::modf(-INFINITY, &f3);
    std::cout << "modf(-Inf) makes " << f3 << " + " << f2 << '\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
modf(-0) makes -0 + -0
modf(-Inf) makes -INF + -0

See also