Modf
Defined in header <cmath>
.
Description
Decomposes given floating point value num
into integral and fractional parts, each having the same type and sign as num
. The integral part (in floating-point format) is stored in the object pointed to by iptr
.
The library provides overloads of std::modf for all cv-unqualified floating-point types as the type of the parameter num
and the pointed-to type of iptr
(od C++23).
Additional Overloads are provided for all integer types, which are treated as double (od C++11).
Declarations
- C++23
- C++11
// 1)
constexpr /* floating-point-type */
modf ( /* floating-point-type */ num,
/* floating-point-type */* iptr );
// 2)
constexpr float modff( float num, float* iptr );
// 3)
constexpr long double modfl( long double num, long double* iptr );
// 4)
template< class Integer >
double modf ( Integer num, double* iptr );
// 1)
float modf ( float num, float* iptr );
// 2)
double modf ( double num, double* iptr );
// 3)
long double modf ( long double num, long double* iptr );
// 4)
float modff( float num, float* iptr );
// 5)
long double modfl( long double num, long double* iptr );
// 6)
template< class Integer >
double modf ( Integer num, double* iptr );
Parameters
num
- floating-point or integer value
iptr
- pointer to floating-point value to store the integral part to
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 Additional Overloads.
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)
.
Examples
#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