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a simple way to print a binary representation of a floating point number?

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Join Date: Oct 2004

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nsavvinv

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a simple way to print a binary representation of a floating point number?

Oct 12th, 2004

Hi everybody,

is there a simple way to print a floating point number in its internal representation
(i.e. binary or hex) in C(C++)? Yet another question -- is there a linux utility to do the opposite thing with a binary file (i.e. choose a range of bits and convert it to a floating point number)?

Thanks.

Best regards,
Nick.

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long time no c

Re: a simple way to print a binary representation of a floating point number?

Oct 12th, 2004

Something like this?

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murschech

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Re: a simple way to print a binary representation of a floating point number?

Jan 31st, 2005

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Originally Posted by nsavvinv

Hi everybody,

is there a simple way to print a floating point number in its internal representation
(i.e. binary or hex) in C(C++)? Yet another question -- is there a linux utility to do the opposite thing with a binary file (i.e. choose a range of bits and convert it to a floating point number)?

Thanks.

Best regards,
Nick.

Here's a program that writes a float or double in the form 2^exp * mantissa, with the mantissa expanded in binary. IEEE specs might call for an offset of the exponent.

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#include <iostream.h>
 
main()
{  double x, mant;  
    int exp = 0;
     cout << "Enter postive x: ";
      //Now find exponent and mantissa
      mant = x;
      while (mant >= 1)
      { mant /= 2;
         exp++;
           // At this point x = mant * 2^(exp)
     }
      while (mant < 0.5) 
      {     mant *= 2;
              exp--;
      }
      cout << x<<" is stored in the computer as \n"<< x <<" = ";
     cout <<"2^(" <<exp<<") *.";
     while (mant > 0)
     {   mant *= 2;
          if (mant >= 1)
         {  cout << '1';
            mant -= 1;
        }
            else
            cout << '0';
    }
          cout << endl;
}
#include <iostream.h>

main()
{  double x, mant;  
    int exp = 0;
     cout << "Enter postive x: ";
      //Now find exponent and mantissa
      mant = x;
      while (mant >= 1)
      { mant /= 2;
         exp++;
           // At this point x = mant * 2^(exp)
     }
      while (mant < 0.5) 
      {     mant *= 2;
              exp--;
      }
      cout << x<<" is stored in the computer as \n"<< x <<" = ";
     cout <<"2^(" <<exp<<") *.";
     while (mant > 0)
     {   mant *= 2;
          if (mant >= 1)
         {  cout << '1';
            mant -= 1;
        }
            else
            cout << '0';
    }
          cout << endl;
}