modf, modff, modfl

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


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


This function behaves as if implemented as follows:

double modf(double value, double *iptr)
    int save_round = fegetround();
    *iptr = std::nearbyint(value);
    return copysign(isinf(value) ? 0.0 : value - (*iptr), value);


#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


  • C11 standard (ISO/IEC 9899:2011):
    • The modf functions (p: 246-247)
    • F.10.3.12 The modf functions (p: 523)
  • C99 standard (ISO/IEC 9899:1999):
    • The modf functions (p: 227)
    • F.9.3.12 The modf functions (p: 460)
  • C89/C90 standard (ISO/IEC 9899:1990):
    • The modf function

See also

rounds to nearest integer not greater in magnitude than the given value
C++ documentation for modf

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