Checking if float is an integer

Question

How can I check if a float variable contains an integer value  So far  I ve been using   float f   4 5886  if  f- int f    0       printf  yes n    else printf  no n      But I wonder if there is a better solution  or if this one has any  or many  drawbacks

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define twop22  0x1 0p 22   define ABS x   fabs x    define isFloatInteger x    ABS x   gt   twop22        ABS x    twop22  - twop22     ABS x

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stdlib float modf  float x  float  ipart  splits into two parts  check if return value  fractional part     0

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if  fmod f  1     0 0              Don t forget math h and libm

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This deals with computational round-off   You set the epsilon as desired  bool IsInteger float value        return fabs ceilf value  - value   lt  EPSILON

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I m not 100  sure but when you cast f to an int  and subtract it from f  I believe it is getting cast back to a float  This probably won t matter in this case  but it could present problems down the line if you are expecting that to be an int for some reason   I don t know if it s a better solution per se  but you could use modulus math instead  for example   float f   4 5886  bool isInt  isInt    f   1 0    0    false   true   depending on your compiler you may or not need the  0 after the 1  again the whole implicit casts thing comes into play  In this code  the bool isInt should be true if the right of the decimal point is all zeroes  and false otherwise

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Keep in mind that most of the techniques here are valid presuming that round-off error due to prior calculations is not a factor  E g  you could use roundf  like this   float z   1 0f   if  roundf z     z        printf  integer n      else       printf  fraction n        The problem with this and other similar techniques  such as ceilf  casting to long  etc   is that  while they work great for whole number constants  they will fail if the number is a result of a calculation that was subject to floating-point round-off error  For example   float z   powf powf 3 0f  0 05f   20 0f    if  roundf z     z        printf  integer n      else       printf  fraction n        Prints  fraction   even though  31 20 20 should equal 3  because the actual calculation result ended up being 2 9999992847442626953125   Any similar method  be it fmodf or whatever  is subject to this  In applications that perform complex or rounding-prone calculations  usually what you want to do is define some  tolerance  value for what constitutes a  whole number   this goes for floating-point equality comparisons in general   We often call this tolerance epsilon  For example  lets say that we ll forgive the computer for up to   - 0 00001 rounding error  Then  if we are testing z  we can choose an epsilon of 0 00001 and do   if  fabsf roundf z  - z   lt   0 00001f        printf  integer n      else       printf  fraction n        You don t really want to use ceilf here because e g  ceilf 1 0000001  is 2 not 1  and ceilf -1 99999999  is -1 not -2   You could use rintf in place of roundf if you prefer   Choose a tolerance value that is appropriate for your application  and yes  sometimes zero tolerance is appropriate   For more information  check out this article on comparing floating-point numbers

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Apart from the fine answers already given  you can also use ceilf f     f or floorf f     f  Both expressions return true if f is an integer  They also returnfalse for NaNs  NaNs always compare unequal  and true for   infinity  and don t have the problem with overflowing the integer type used to hold the truncated result  because floorf   ceilf   return floats

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if  f  lt   LONG MIN    f  gt   LONG MAX    f     long f     it s an integer

[c] Checking if float is an integer

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