How to extract the decimal part from a floating point number in C

Question

How can we extract the decimal part of a floating point number and store the decimal part and the integer part into two separate integer variables

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Maybe the best idea is to solve the problem while the data is in String format  If you have the data as String  you may parse it according to the decimal point  You extract the integral and decimal part as Substrings and then convert these substrings to actual integers

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The quick  in a nut shell  most obvious answer seems like    define N DECIMAL POINTS PRECISION  1000     n   3  Three decimal points   float f   123 456  int integerPart    int f  int decimalPart     int  f N DECIMAL POINTS PRECISION  N DECIMAL POINTS PRECISION     You would change how many decimal points you want by changing the N DECIMAL POINTS PRECISION to suit your needs

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I think that using string is the correct way to go in this case  since you don t know a priori the number of digits in the decimal part  But  it won t work for all cases  e g  1 005   as mentioned before by  SingleNegationElimination  Here is my take on this    include  lt stdio h gt   include  lt stdlib h gt   int main void        char s value 60   s integral 60   s fractional 60       int i  found   0  count   1  integral  fractional       scanf   s   s value        for  i   0  s value i       0   i                  if   found                        if  s value i                                        found   1                  s integral i      0                   continue                            s integral i    s value i               count                      else             s fractional i - count    s value i             s fractional i - count      0        integral   atoi s integral       fractional   atoi s fractional       printf  value    s  integral    d  fractional    d n           s value  integral  fractional        return 0

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cout lt  lt  enter a decimal number n   cin gt  gt str  for i 0 i lt str size   i          if str i            break     for j i 1 j lt str size   j          cout lt  lt str j

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I made this function  it seems to work fine    include  lt math h gt   void GetFloattoInt  double fnum  long precision  long  pe  long  pd      long pe sign    long intpart    float decpart     if fnum gt  0          pe sign 1        else         pe sign -1         intpart  long fnum    decpart fnum-intpart      pe intpart       pd    long  decpart pe sign pow 10 precision     long pow 10 precision

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Try this   int main       double num   23 345    int intpart    int num    double decpart   num - intpart    printf  Num    f  intpart    d  decpart    f n   num  intpart  decpart       For me  it produces   Num   23 345000  intpart   23  decpart   0 345000   Which appears to be what you re asking for

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You use the modf function   double integral  double fractional   modf some double   amp integral     You can also cast it to an integer  but be warned you may overflow the integer  The result is not predictable then

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Use the floating number to subtract the floored value to get its fractional part   double fractional   some double - floor some double     This prevents the typecasting to an integer  which may cause overflow if the floating number is very large that an integer value could not even contain it   Also for negative values  this code gives you the positive fractional part of the floating number since floor   computes the largest integer value not greater than the input value

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Suppose A  is your integer then  int A  means casting the number to an integer and will be the integer part  the other is  A -  int A  10 n  here n is the number of decimals to keep

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Even I was thinking how to do it  But I found a way  Try this code  printf  Enter a floating number    scanf   d c d    amp no   amp dot   amp dec   printf  Number  d Decimal part  d   no  dec     Output -  Enter a floating number 23 13 Number 23 Decimal part 13

User · Answer

Use the floating number to subtract the floored value to get its fractional part   double fractional   some double - floor some double     This prevents the typecasting to an integer  which may cause overflow if the floating number is very large that an integer value could not even contain it   Also for negative values  this code gives you the positive fractional part of the floating number since floor   computes the largest integer value not greater than the input value

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Try this   int main       double num   23 345    int intpart    int num    double decpart   num - intpart    printf  Num    f  intpart    d  decpart    f n   num  intpart  decpart       For me  it produces   Num   23 345000  intpart   23  decpart   0 345000   Which appears to be what you re asking for

User · Answer

You use the modf function   double integral  double fractional   modf some double   amp integral     You can also cast it to an integer  but be warned you may overflow the integer  The result is not predictable then

User · Answer

Try this   int main       double num   23 345    int intpart    int num    double decpart   num - intpart    printf  Num    f  intpart    d  decpart    f n   num  intpart  decpart       For me  it produces   Num   23 345000  intpart   23  decpart   0 345000   Which appears to be what you re asking for

User · Answer

You use the modf function   double integral  double fractional   modf some double   amp integral     You can also cast it to an integer  but be warned you may overflow the integer  The result is not predictable then

User · Answer

include  lt stdio h gt  Int main      float f 56 75  int a  int f  int result  f-a  100  printf   integer    d n decimal part to integer     d n  result     Output -  integer  56 decimal part to integer   75

User · Answer

Maybe the best idea is to solve the problem while the data is in String format  If you have the data as String  you may parse it according to the decimal point  You extract the integral and decimal part as Substrings and then convert these substrings to actual integers

User · Answer

Here is another way    include  lt stdlib h gt  int main         char  inStr    123 4567             the number we want to convert     char  endptr                        unused char ptr for strtod      char  loc   strchr inStr            long mantissa   strtod loc 1  endptr       long whole   strtod inStr  endptr        printf  whole   d  n   whole         whole number portion     printf  mantissa   d   mantissa      decimal portion      http   codepad org jyHoBALU  Output   whole  123  mantissa  4567

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The quick  in a nut shell  most obvious answer seems like    define N DECIMAL POINTS PRECISION  1000     n   3  Three decimal points   float f   123 456  int integerPart    int f  int decimalPart     int  f N DECIMAL POINTS PRECISION  N DECIMAL POINTS PRECISION     You would change how many decimal points you want by changing the N DECIMAL POINTS PRECISION to suit your needs

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cout lt  lt  enter a decimal number n   cin gt  gt str  for i 0 i lt str size   i          if str i            break     for j i 1 j lt str size   j          cout lt  lt str j

User · Answer

You use the modf function   double integral  double fractional   modf some double   amp integral     You can also cast it to an integer  but be warned you may overflow the integer  The result is not predictable then

User · Answer

I think that using string is the correct way to go in this case  since you don t know a priori the number of digits in the decimal part  But  it won t work for all cases  e g  1 005   as mentioned before by  SingleNegationElimination  Here is my take on this    include  lt stdio h gt   include  lt stdlib h gt   int main void        char s value 60   s integral 60   s fractional 60       int i  found   0  count   1  integral  fractional       scanf   s   s value        for  i   0  s value i       0   i                  if   found                        if  s value i                                        found   1                  s integral i      0                   continue                            s integral i    s value i               count                      else             s fractional i - count    s value i             s fractional i - count      0        integral   atoi s integral       fractional   atoi s fractional       printf  value    s  integral    d  fractional    d n           s value  integral  fractional        return 0

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Here is another way    include  lt stdlib h gt  int main         char  inStr    123 4567             the number we want to convert     char  endptr                        unused char ptr for strtod      char  loc   strchr inStr            long mantissa   strtod loc 1  endptr       long whole   strtod inStr  endptr        printf  whole   d  n   whole         whole number portion     printf  mantissa   d   mantissa      decimal portion      http   codepad org jyHoBALU  Output   whole  123  mantissa  4567

User · Answer

Even I was thinking how to do it  But I found a way  Try this code  printf  Enter a floating number    scanf   d c d    amp no   amp dot   amp dec   printf  Number  d Decimal part  d   no  dec     Output -  Enter a floating number 23 13 Number 23 Decimal part 13

User · Answer

I made this function  it seems to work fine    include  lt math h gt   void GetFloattoInt  double fnum  long precision  long  pe  long  pd      long pe sign    long intpart    float decpart     if fnum gt  0          pe sign 1        else         pe sign -1         intpart  long fnum    decpart fnum-intpart      pe intpart       pd    long  decpart pe sign pow 10 precision     long pow 10 precision

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If you just want to get the first decimal value  the solution is really simple   Here s an explanatory example   int leftSideOfDecimalPoint    int  initialFloatValue     The cast from float to int keeps only the integer part  int temp    int  initialFloatValue   10  int rightSideOfDecimalPoint   temp   10    Say for example we have an initial float value of 27 8      By just casting the initial float value to an int  you discard the fraction part and only keep the integer part  By multiplying the initial float value by 10 we get a result of 278 0  then by casting this result to int  gives you the value of 278 If we divide 278 by 10  we get 27 8  of which the remainder is 8  which is the value at the right side of the decimal point  Thus use modulus    This technique can then be used to get the following decimal characters by using for example 100 instead of 10  and so on     Just take note that if you use this technique on real-time systems  for example to display it on a 7-segment display  it may not work properly because we are multiplying with a float value  where multiplication takes a lot of overhead time

User · Answer

Suppose A  is your integer then  int A  means casting the number to an integer and will be the integer part  the other is  A -  int A  10 n  here n is the number of decimals to keep

User · Answer

I created a subroutine one using a double float  it returns 2 integer values    void double2Ints double f  int p  int  i  int  d          f   float  p decimal precision  i integer  d decimal   int   li     int   prec 1     for int x p x 0 x--           prec  10           same as power 10 p     li    int  f                  get integer part    d    int    f-li  prec       get decimal part    i   li     void test        double df   3 14159265    int   i d    for int p 2 p lt 9 p            double2Ints df  p   i  d   printf  d2i   d   f    d  d r n  p  df i d

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Try this   int main       double num   23 345    int intpart    int num    double decpart   num - intpart    printf  Num    f  intpart    d  decpart    f n   num  intpart  decpart       For me  it produces   Num   23 345000  intpart   23  decpart   0 345000   Which appears to be what you re asking for

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include  lt stdio h gt  Int main      float f 56 75  int a  int f  int result  f-a  100  printf   integer    d n decimal part to integer     d n  result     Output -  integer  56 decimal part to integer   75

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I created a subroutine one using a double float  it returns 2 integer values    void double2Ints double f  int p  int  i  int  d          f   float  p decimal precision  i integer  d decimal   int   li     int   prec 1     for int x p x 0 x--           prec  10           same as power 10 p     li    int  f                  get integer part    d    int    f-li  prec       get decimal part    i   li     void test        double df   3 14159265    int   i d    for int p 2 p lt 9 p            double2Ints df  p   i  d   printf  d2i   d   f    d  d r n  p  df i d

User · Answer

If you just want to get the first decimal value  the solution is really simple   Here s an explanatory example   int leftSideOfDecimalPoint    int  initialFloatValue     The cast from float to int keeps only the integer part  int temp    int  initialFloatValue   10  int rightSideOfDecimalPoint   temp   10    Say for example we have an initial float value of 27 8      By just casting the initial float value to an int  you discard the fraction part and only keep the integer part  By multiplying the initial float value by 10 we get a result of 278 0  then by casting this result to int  gives you the value of 278 If we divide 278 by 10  we get 27 8  of which the remainder is 8  which is the value at the right side of the decimal point  Thus use modulus    This technique can then be used to get the following decimal characters by using for example 100 instead of 10  and so on     Just take note that if you use this technique on real-time systems  for example to display it on a 7-segment display  it may not work properly because we are multiplying with a float value  where multiplication takes a lot of overhead time

[c] How to extract the decimal part from a floating point number in C?

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