Why is (double)0.6f > (double)(6/10f)?

Question

This is what happens on my computer:

(double)(float)0.6
= 0.60000002384185791

(double)0.6f
= 0.60000002384185791

(double)(6/10f)
= 0.6

(double)(float)(6/1         


        

Answer 1

If I were a betting man, I'd say the difference is in where the coercion is happening. In the latter two examples (the ones with 6/10f), there are two literals that are both whole numbers (the integer 6 and the float 10.00000000...). The division appears to be happening after the coercion, at least in the compiler you're using. In the first two examples, you have a fractional float literal (0.6) which cannot be adequately expressed as a binary value within the mantissa of a float. Coercing that value to a double cannot repair the damage that was already done.

In the environments that are producing completely consistent results, the division is occurring before the coercion to double (the 6 will be coerced to a float for the division to match the 10, the division is carried out in float space, then the result is coerced to a double).