From 82ba818ff456bcd6d56a06226e3f27e98fbb55c3 Mon Sep 17 00:00:00 2001 From: Craig Jennings Date: Thu, 14 Aug 2025 22:58:58 -0500 Subject: removing all downloaded devdocs files --- devdocs/c/numeric%2Fmath%2Fmodf.html | 78 ------------------------------------ 1 file changed, 78 deletions(-) delete mode 100644 devdocs/c/numeric%2Fmath%2Fmodf.html (limited to 'devdocs/c/numeric%2Fmath%2Fmodf.html') diff --git a/devdocs/c/numeric%2Fmath%2Fmodf.html b/devdocs/c/numeric%2Fmath%2Fmodf.html deleted file mode 100644 index 7ee360cd..00000000 --- a/devdocs/c/numeric%2Fmath%2Fmodf.html +++ /dev/null @@ -1,78 +0,0 @@ -

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),

-

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

See also

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

- © cppreference.com
Licensed under the Creative Commons Attribution-ShareAlike Unported License v3.0.
- https://en.cppreference.com/w/c/numeric/math/modf -

-
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