Question

Can every possible value of a float variable can be represented exactly in a double variable?

In other words, for all possible values X will the following be successful:

float f1 = X; double d = f1; float f2 = (float)d;  if(f1 == f2)   System.out.println('Success!'); else   System.out.println('Failure!'); 

My suspicion is that there is no exception, or if there is it is only for an edge case (like +/- infinity or NaN).

Edit: Original wording of question was confusing (stated two ways, one which would be answered ‘no’ the other would be answered ‘yes’ for the same answer). I’ve reworded it so that it matches the question title.

Answer 1

Yes.

Proof by enumeration of all possible cases:

public class TestDoubleFloat  {     public static void main(String[] args) {         for (long i = Integer.MIN_VALUE; i <= Integer.MAX_VALUE; i++) {             float f1 = Float.intBitsToFloat((int) i);             double d = (double) f1;             float f2 = (float) d;             if (f1 != f2) {                 if (Float.isNaN(f1) && Float.isNaN(f2)) {                     continue; // ok, NaN                 }                 fail('oops: ' + f1 + ' != ' + f2);             }         }     } } 

finishes in 12 seconds on my machine. 32 bits are small.