Can 12.1 be represented exactly as a floating point number?

Question

This is in reference to the comments in this question:

This code in Java produces 12.100000000000001 and this is using 64-bit doubles which can present 12.1 exactly. – Pyrolistical

Is this true? I felt that since a floating point number is represented as a sum of powers of two, you cannot represent 12.1 exactly, no matter how many bits you have. However, when I implemented both the algorithms and printed the results of calling them with (12.1, 3) with many significant digits, I get, for his and mine respectively:

12.10000000000000000000000000000000000000000000000000000000000000000000000000 12.10000000000000100000000000000000000000000000000000000000000000000000000000

I printed this using String.format("%76f"). I know that's more zeros than necessary, but I don't see any rounding in the 12.1 .

Answer 1

No. As others noted in followups to his comment, no sum of (a finite number of) powers of two can ever add up to exactly 12.1. Just like you can't represent 1/3 exactly in base ten, no matter how many digits you use after the decimal point.

Answer 2

In binary, 12.1 is:

1100.000110011001100110011...

Since this doesn't terminate, it can't be represented exactly in the 53 significand bits of a double, or any other finite-width binary floating-point type.