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modf, modff, modfl

Defined in header `<math.h>`
float modff( float arg, float* iptr );	(1)	(since C99)
double modf( double arg, double* iptr );	(2)
long double modfl( long double arg, long double* iptr );	(3)	(since C99)

1-3) Decomposes given floating point value arg into integral and fractional parts, each having the same type and sign as arg. The integral part (in floating-point format) is stored in the object pointed to by iptr.

Parameters

arg	-	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 arg (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 arg is ±0, ±0 is returned, and ±0 is stored in *iptr.
If arg is ±∞, ±0 is returned, and ±∞ is stored in *iptr.
If arg 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 value, double *iptr)
{
#pragma STDC FENV_ACCESS ON
    int save_round = fegetround();
    fesetround(FE_TOWARDZERO);
    *iptr = std::nearbyint(value);
    fesetround(save_round);
    return copysign(isinf(value) ? 0.0 : value - (*iptr), value);
}

Example

#include <stdio.h>
#include <math.h>
#include <float.h>
 
int main(void)
{
    double f = 123.45;
    printf("Given the number %.2f or %a in hex,\n", f, f);
 
    double f3;
    double f2 = modf(f, &f3);
    printf("modf() makes %.2f + %.2f\n", f3, f2);
 
    int i;
    f2 = frexp(f, &i);
    printf("frexp() makes %f * 2^%d\n", f2, i);
 
    i = ilogb(f);
    printf("logb()/ilogb() make %f * %d^%d\n", f/scalbn(1.0, i), FLT_RADIX, i);
 
    // special values
    f2 = modf(-0.0, &f3);
    printf("modf(-0) makes %.2f + %.2f\n", f3, f2);
    f2 = modf(-INFINITY, &f3);
    printf("modf(-Inf) makes %.2f + %.2f\n", f3, f2);
}

Possible output:

Given the number 123.45 or 0x1.edccccccccccdp+6 in hex,
modf() makes 123.00 + 0.45
frexp() makes 0.964453 * 2^7
logb()/ilogb() make 1.92891 * 2^6
modf(-0) makes -0.00 + -0.00
modf(-Inf) makes -INF + -0.00

References

C11 standard (ISO/IEC 9899:2011):

7.12.6.12 The modf functions (p: 246-247)
F.10.3.12 The modf functions (p: 523)

C99 standard (ISO/IEC 9899:1999):

7.12.6.12 The modf functions (p: 227)
F.9.3.12 The modf functions (p: 460)

C89/C90 standard (ISO/IEC 9899:1990):

4.5.4.6 The modf function

trunctruncftruncl (C99)(C99)(C99)	rounds to nearest integer not greater in magnitude than the given value (function)
C++ documentation for `modf`