std::modf,std::modff,std::modfl(3)

NAME
std::modf,std::modff,std::modfl - std::modf,std::modff,std::modfl

Synopsis
   Defined in header <cmath>
   float modf ( float x, float* iptr ); (1) (constexpr since C++23)
   float modff( float x, float* iptr ); (2) (since C++11)
                        (constexpr since C++23)
   double modf ( double x, double* iptr ); (3) (constexpr since C++23)
   long double modf ( long double x, long double* iptr ); (4) (constexpr since C++23)
   long double modfl( long double x, long double* iptr ); (5) (since C++11)
                        (constexpr since C++23)
   1-5) Decomposes given floating point value x into integral and fractional parts,
   each having the same type and sign as x. The integral part (in floating-point
   format) is stored in the object pointed to by iptr.

Parameters
   x - floating point value
   iptr - pointer to floating point value to store the integral part to

Return value
   If no errors occur, returns the fractional part of x with the same sign as x. 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 x (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 x is ±0, ±0 is returned, and ±0 is stored in *iptr.
     * If x is ±∞, ±0 is returned, and ±∞ is stored in *iptr.
     * If x 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 x, double* iptr)
 {
 #pragma STDC FENV_ACCESS ON
     int save_round = std::fegetround();
     std::fesetround(FE_TOWARDZERO);
     *iptr = std::nearbyint(x);
     std::fesetround(save_round);
     return std::copysign(std::isinf(x) ? 0.0 : x - (*iptr), x);
 }

Example
   Compares different floating-point decomposition functions // Run this code
 #include <iostream>
 #include <cmath>
 #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';
 }

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
 modf(-0) makes -0 + -0
 modf(-Inf) makes -INF + -0

See also
   trunc
   truncf
   truncl nearest integer not greater in magnitude than the given value
   (C++11) (function)
   (C++11)
   (C++11)