modf, modff, modfl
From cppreference.com
| 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.Contents |
[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 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).
[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
argis ±0, ±0 is returned, and ±0 is stored in *iptr. - If
argis ±∞, ±0 is returned, and ±∞ is stored in *iptr. - If
argis 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 <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
[edit] 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
[edit] See also
| (C99)(C99)(C99) |
rounds to nearest integer not greater in magnitude than the given value (function) |
| C++ documentation for modf
| |
