hello friends, Help! Please help me out, what is the process of solving this problem, what should I do? Using the floating point format convert the floating point number represented by 11000100110101101111010011010000 to base 10 (decimal) representation where, Sign bit = negative = 1 Base 2 8-bit excess-128 exponent 23 bits of mantissa Thanks.. Manoj Kumar Paul Software Developer Java

I am going to assume that is IEEE 754 format. And that you are working in Java, at least 1.5. So I think it is pretty easy: long along = Long.parseLong("11000100110101101111010011010000",2); float afloat = Float.intBitsToFloat( (int)along ); System.out.println("along= " + along); System.out.println("afloat= " + afloat); You might also take a look at: http://en.wikipedia.org/wiki/IEEE_754

Basically, you do what we did for the floating point numbers, only backwards: Step 0: 1 10001001 10101101111010011010000 becomes: Sign = 1 e = 10001001 s = 10101101111010011010000 Answer = ? Step 1: Convert the 'e' to base-10, add the "1." to the front of the 's'. e = 137 - 127 (b10) s = 1.1010110111101001101 Answer = -? Step 2: e = e - 127 e = 10 (b10) s = 1.1010110111101001101 Answer = -? Step 3: Move the '.' e number of spaces. s = 11010110111.101001101 Answer = -? Step 4: Our answer (in base-2) Answer = -11010110111.101001101(b2) Step 5: base-2 -> base-10 Answer = -(1719 + 2^-1 + 2^-3 + 2^-6 + 2^-7 + 2^-9) = -1719.650390625

Is -1719.650390625 the final answer to the question? Manoj Kumar Paul Software Developer Java

Thank you Razor, Is -1719.650390625 the final answer to the question? Thanx once again. Manoj Kumar Paul Software Developer Java

Well, VBA says so, so that's what I'm going with.

Ok, Thanx a loads Razor Manoj Kumar Paul Software Developer Java