std::modf
From cppreference.com
Defined in header
<cmath>
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float modf( float x, float* iptr );
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(1) | |
double modf( double x, double* iptr );
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(2) | |
long double modf( long double x, long double* iptr );
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(3) | |
1-3) Decomposes given floating point value
x
into integral and fractional parts, each having the same type and sign as x
. The integral part (in floating-point format) is stored in the object pointed to by iptr
.
Contents |
[edit] Parameters
x | - | 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 x
(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
x
is ±0, ±0 is returned, and ±0 is stored in *iptr. - If
x
is ±∞, ±0 is returned, and ±∞ is stored in *iptr. - If
x
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 x, double* iptr) { #pragma STDC FENV_ACCESS ON int save_round = std::fegetround(); std::fesetround(FE_TOWARDZERO); *iptr = std::nearbyint(value); std::fesetround(save_round); return std::copysign(std::isinf(x) ? 0.0 : x - (*iptr), x); }
[edit] Example
Compares different floating-point decomposition functions
Run this code
#include <iostream> #include <cmath> #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'; }
Possible output:
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
[edit] See also
(C++11)
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nearest integer not greater in magnitude than the given value (function) |
C documentation for modf
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