CODEWITHSUNDEEP
c++stlfloating-pointdoubleprecision
user277465
user277465

Reputation:

C++ code to convert a fraction to binary

I just wrote the following C++ code in order to convert a fractional number into its corresponding binary format.

double get_decimal_part(double num) {
   long x = static_cast<long>(num);
   return (num - static_cast<double>(x));
}

long get_real_part(double num) {
   return static_cast<long>(num);
}

string fraction_to_binary(double num) {
   string decimal_binary = "";
   double decimal_part = get_decimal_part(num);
   while ( decimal_part > 0 ) {
      double temp = decimal_part * 2;
      if ( get_real_part(temp) == 0 ) decimal_binary += "0";
      else                            decimal_binary += "1";
      decimal_part = get_decimal_part(temp); 
   }
   return decimal_binary;
}

int main() {
   cout << "3.50 - " << fraction_to_binary(3.50) << endl;
   cout << "3.14 - " << fraction_to_binary(3.14) << endl;
}

The output would be :-

3.50 - 1
3.14 - 001000111101011100001010001111010111000010100011111

I'd have the following questions regarding the same :-

  1. In the case if "3.50", my implementation would give "1" as the output -- how can I go about modifying my implementation in order to account for the trailing "0" in 3.50?
  2. If there were any library functions that could help me get the precision of a floating point number? I'm guessing I could use that information to modify my implementation.

[EDIT] I also tried using the following to convert a float to a string but it wouldnt help either.

   stringstream ss;
   ss << my_float;
   cout << string(ss.str()) << endl;

Upvotes: 0

Views: 2932

Answers (1)

James Kanze
James Kanze

Reputation: 154007

Before answering your specific questions, what's wrong with modf for this?

With regards to your specific questions:

  1. What trailing "0"? You're talking about a text representation here. Inside the machine, "3.5" and "3.50" correspond to the same number, and have the same representation.

  2. There is a library function which returns the precision of a floating point number: std::numeric_limits<double>::digits (except that it isn't a function, but a constant). But if you want to break a number down into its integral and whole number parts, modf fits the bill exactly. And unlike your code, will actually work, for all values of double.

EDIT:

Looking closer at the larger picture of what you are trying to do: my approach would be to use frexp to extract the base 2 exponent, then ldexp to scale the number into the range [0.5...1) Then loop std::numeric_limits<double>::digits times, each time multiplying by 2, and checking that: if the results of the multiplication are less than 1, then insert a 0 digit; otherwise, insert a 1 digit and subtract 1. (Note that all of the above actions will be exact if the machine floating point is base 2, or a power of 2.)

Upvotes: 3

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