printing the integral part of a floating point number

Question

I am trying to figure out how to print floating point numbers without using library functions. Printing the decimal part of a floating point number turned out to be quite ea

Answer 1

According to IEEE single precision float implementation, only 24 bits of data is stored at any time in a float variable. This means only maximum 7 decimal digits are stored in the floating number.

Rest of the hugeness of the number is stored in the exponent. FLT_MAX is initialized as 3.402823466e+38F. So, after the 10th precision, which digit should get printed is not defined anywhere.

From Visual C++ 2010 compiler, I get this output 340282346638528860000000000000000000000.000000, which is the only vaild output.

So, initially we have these many valid digits 3402823466 So after the 1st division we have only 0402823466 So, the system need to get rid of the left 0 and introduce a new digit at the right. In ideal integer division, it is 0. Because you are doing floating division (value /= base;) , system is getting some other digit to fill in that location.

So, in my opinion, the printf could be assigning the above available significant digits to an integer and working with this.