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