Conversion of hexadecimal to floating point

Question

Does anybody know any way of conversion of hexadecimal to floating point? Any C, Java, C++ or any language's program that could automatically convert it or the formula to convert it manually on page and copy.

Any help would be highly appreciated.

Answer 1

In C and C++, the direct conversion goes as follows

int i = 0xFF2;
float f = 0xFF2.25p10;
cout<<"i = "<<i<<endl<<"f = "<<f;

Gives output as

i = 4082
f = 4.18012e+06

However, this does not highlight the logic behind the conversion, the same can be carried out on pen and paper as follows:

1. Write the whole hexadecimal string on a paper.
2. Write indices below each character of the string. Starting from period (.) and going leftwards, assign index 0, 1, 2, and so on. Starting from period and going rightwards, assign index -1, -2, -3, and so on.

3. Using the following mapping, convert each of the characters into their equivalent decimal notation before multiplying in step 4.

   

4. If A is the string that represents your hexadecimal number, then

   decimal_equivalent = ∑ A[i]*(16ⁱ)
   where, i is the index in A.

For example, consider the number FF2.25,

 ----------------------------------------------------
|    F    |    F    |    2     .    2     |    5     | <--- hex number
-----------------------------------------------------
|   2     |    1    |    0     |    -1    |   -2     | <--- index
 ----------------------------------------------------
|  16**2  |  16**1  |  16**0   | 16**(-1) | 16**(-2) | <--- 16**index
 ----------------------------------------------------
|  F*256  |  F*16   |   2*1    | 2*0.0625 | 5*0.0039 | <--- A[index]*(16**(index))
 ----------------------------------------------------
|  15*256 |  15*16  |   2*1    | 2*0.0625 | 5*0.0039 | <--- conversion
 ----------------------------------------------------
|   3840  +   240   +    2     +   0.125  +  0.0195  =  4082.1445
 ----------------------------------------------------  

Consider the below C++ code snippet as a pseudocode, I haven't tested it(sorry for that).

int indexWithPeriod(string s)
{
     // returns index at which '.' is stored; else -1
     int len = s.length()-1;
     while (len--)
     {
          if (s[len] == '.')
             return len;
     }
     return -1;
}
double hexToDec(string hex)
{
     // returns decimal equivalent of a hexa decimal number stored in string
     double ans = 0;
     int len = hex.length();
     int periodAt = indexWithPeriod(hex);
     int i = periodAt-1, k = 0;
     while (i >= 0)
     {
           ans += hex[i]*pow(16, k);
           --i;
           ++k;
     }
     i = periodAt+1;
     k = -1;
     while (i < len)
     {
           ans += hex[i]*pow(16, k)
           ++i;
           --k;
     }
     return ans;
}

The function hexToDec() will convert the given hexadecimal number in string hex to a double precision floating point number(returning its value).

There is an implicit pow() function in cmath, and a length() function in cstring.