Comparing float and double

Solution 1:

This is because 1.1 is not exactly representable in binary floating-point. But 1.5 is.

As a result, the float and double representations will hold slightly different values of 1.1.

Here is exactly the difference when written out as binary floating-point:

(float) 1.1 = (0.00011001100110011001101)₂
(double)1.1 = (0.0001100110011001100110011001100110011001100110011010)₂

Thus, when you compare them (and the float version gets promoted), they will not be equal.

Solution 2:

Must read: What Every Computer Scientist Should Know About Floating-Point Arithmetic

Solution 3:

The exact value of 1.1 decimal in binary is non-ending fraction 1.00011001100110011001100(1100).... The double constant 1.1 is 53-bit truncation / approximate value of that mantissa. Now this when converted to float, the mantissa will be represented just in 24 bits.

When the float is converted back to double, the mantissa is now back to 53 bits, but all memory of the digits beyond 24 are lost - the value is zero-extended, and now you're comparing (for example, depending on the rounding behaviour)

1.0001100110011001100110011001100110011001100110011001

and

1.0001100110011001100110000000000000000000000000000000

Now, if you used 1.5 instead of 1.1;

1.5 decimal is exactly 1.1 in binary. It can be presented exactly in just 2 bit mantissa, therefore even the 24 bits of float are an exaggeration... what you have is

1.1000000000000000000000000000000000000000000000000000

and

1.10000000000000000000000

The latter, zero extended to a double would be

1.1000000000000000000000000000000000000000000000000000

which clearly is the same number.