如何从C中的浮点数中提取小数部分？

Question

我们如何提取浮点数的小数部分并将小数部分和整数部分存储到两个单独的整数变量中？

Answer 1

You use the modf function:

double integral;
double fractional = modf(some_double, &integral);

You can also cast it to an integer, but be warned you may overflow the integer. The result is not predictable then.

Answer 2

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.

Answer 3

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.

Answer 4

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 (eg 1.005), as mentioned before by @SingleNegationElimination. Here is my take on this:

#include <stdio.h>
#include <stdlib.h>

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;
}

Answer 5

Use the floating number to subtract the floor ed 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 .

Answer 6

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<9;p++)
  {
    double2Ints(df, p, &i,&d); printf("d2i (%d) %f = %d.%d\r\n",p, df,i,d);
  }
}

Answer 7

Here is another way:

#include <stdlib.h>
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

Answer 8

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", &no, &dot, &dec);
printf("Number=%d Decimal part=%d", no, dec);

Output:-

Enter a floating number
23.13
Number=23 Decimal part=13

Answer 9

I made this function, it seems to work fine:

#include <math.h>

void GetFloattoInt (double fnum, long precision, long *pe, long *pd)
{
  long pe_sign;
  long intpart;
  float decpart;

  if(fnum>=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));
}

Answer 10

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.

Answer 11

 #include <stdio.h>
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

Answer 12

假设 A 是您的整数，则 (int)A，表示将数字转换为整数并将是整数部分，另一个是 (A - (int)A)*10^n，这里 n 是要保留的小数位数.

Answer 13

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.

Answer 14

float num;
int intgPart;
float fracPart;
printf("Enter the positive floating point number: ");
scanf("%f", &num);

intgPart = (int)num;
fracPart = num - intgPart;

The fractional part can be obtained by subtracting an integral part from the original double value.

Answer 15

cout<<"enter a decimal number\n";
cin>>str;
for(i=0;i<str.size();i++)
{
    if(str[i]=='.')
    break;
}

for(j=i+1;j<str.size();j++)
{
    cout<<str[j];
}