[edit]Parameters

[edit]Return value

If no errors occur, returns the fractional part ofarg with the same sign asarg. The integral part is put into the value pointed to byiptr.

The sum of the returned value and the value stored in*iptr givesarg (allowing for rounding).

[edit]Error handling

This function is not subject to any errors specified inmath_errhandling.

If the implementation supports IEEE floating-point arithmetic (IEC 60559),

• Ifarg is ±0, ±0 is returned, and ±0 is stored in*iptr.
• Ifarg is ±∞, ±0 is returned, and ±∞ is stored in*iptr.
• Ifarg 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 ONint save_round=fegetround();fesetround(FE_TOWARDZERO);*iptr= std::nearbyint(value);fesetround(save_round);returncopysign(isinf(value)?0.0: value-(*iptr), value);}

[edit]Example

#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);}

Given the number 123.45 or 0x1.edccccccccccdp+6 in hex,modf() makes 123.00 + 0.45frexp() makes 0.964453 * 2^7logb()/ilogb() make 1.92891 * 2^6modf(-0) makes -0.00 + -0.00modf(-Inf) makes -INF + -0.00

[edit]References

[edit]See also

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