Convert the array to a number starting with 0.x (C) - c

Question

I have a function for translating the fractional part of the entered number into another number system (not the most beautiful code):
void FracToAny(double D, int q, int t)
{
double b[10] = {0};
for (int i = 1; i < t; i++)
{
double DP = D * q;
D = modf(DP, &b[i]);
if (D == 0)
{
break;
}
D = DP - b[i];
}
for (int i = 0; i < t; i++)
{
if (q == 16)
{
switch ((int)b[i])
{
case 10:
printf("A");
break;
case 11:
printf("B");
break;
case 12:
printf("C");
break;
case 13:
printf("D");
break;
case 14:
printf("E");
break;
case 15:
printf("F");
break;
default:
printf("%d", (int)b[i]);
break;
}
}
}
}
As a result, I get an array of double, which is a character-by-character representation of the resulting fractional part. However, I would like to return this value from the function in the form of "0.result". How can I do this?
It would be good not to use union or dynamic memory allocation (I've seen them trying to find a solution to the problem). I'd like to make it as simple as possible.

Answer 1

I think this is what you are trying to do (see my comment above):
Mac_3.2.57$cat etFract.c
#include <stdio.h>
#include <math.h>
int main(void)
{
int i;
double D;
int t = 3;
double b[10] = {0};
D = (3 * 16*16*16 + 4 * 16*16 + 5 *16)/(16*16*16.0);
printf("translating %f...\n", D);
for(i = 0; i < t; i++){
b[i] = modf(D, &D);
if(D == 0){
break;
}
D = b[i] * 16;
}
printf("0.");
for(int i = 0; i < t; i++){
switch((int)b[i]*16){
case 10:
printf("A\n");
break;
case 11:
printf("B\n");
break;
case 12:
printf("C\n");
break;
case 13:
printf("D\n");
break;
case 14:
printf("E\n");
break;
case 15:
printf("F\n");
break;
default:
printf("%d", (int)(b[i]*16));
break;
}
}
printf("\n");
return(0);
}
Mac_3.2.57$cc etFract.c
Mac_3.2.57$./a.out
translating 3.269531...
0.450
Mac_3.2.57$