How to form a Double with an Integer and Fractional parts

Question

I am given two values, the integer and the fractional parts of a double. What is the best way to form the double value? Essentially, it would be the opposite of modf().

I'm doing this in swift, but any mathematical way is what I'm looking for.

let integer = 12
let fractional = 34
let d = ???(12, 34) // 12.34

Answer 1

For general case you have to find the number of digits are there in the fraction part. So I am assuming that x as integer part and y as fractional part. Then you have to divide y by 10 to the power of number of digits in y. And then you have to add x to the updated y value.

Here is the code,

#include <stdio.h>
#include<math.h>
int main()
{
    int x,y;
    scanf("%d %d",&x,&y);
    int count = 0;
    int z = y;
    while(z>0)
    {
        count+=1;
        z = z/10;
    }
    float ans = 0;
    ans = (double)x + (double)y/pow(10,count);
    printf("%f",ans);
    return 0;
}

Hope this solves your problem.

If x and y are also considered as float there is even a better solution. Thanks to Joël Hecht

#include <stdio.h>
#include<math.h>
int main()
{
    double x,y;
    scanf("%lf %lf",&x,&y);
    while(y>=1)
    {
        y = y/10.0;
    }
    printf("%lf",x+y);
    return 0;
}

P.S :- You have to take care if the number given is negative.