modf, modff, modfl
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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.
[edit] Parameters
arg
-
floating-point value
iptr
-
pointer to floating-point value to store the integral part to
[edit] Return value
If no errors occur, returns the fractional part of arg with the same sign as arg. 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).
[edit] 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.
[edit] 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); }
[edit] Example
Run this code
#include <float.h> #include <math.h> #include <stdio.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
[edit] References
- C23 standard (ISO/IEC 9899:2024):
- 7.12.6.12 The modf functions (p: TBD)
- F.10.3.12 The modf functions (p: TBD)
- C17 standard (ISO/IEC 9899:2018):
- 7.12.6.12 The modf functions (p: TBD)
- F.10.3.12 The modf functions (p: TBD)
- 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
[edit] See also
(C99)(C99)(C99)
(function) [edit]
C++ documentation for modf