Why when I assign an int to a float is it not the same value?
java
Solution
A `float` has the same number of bits as an `int` -- 32 bits. But it allows for a greater range of values, far greater than the range of `int` values. However, the precision is fixed at 24 bits (23 "mantissa" bits, plus 1 implied `1` bit). At the value of about 63,000,000, the precision is greater than `1`. You can verify this with the `Math.ulp` method, which gives the difference between 2 consecutive values.
The code
System.out.println(Math.ulp(63000000.0f));
prints
4.0
You can use `double` values for a far greater (yet still limited) precision:
System.out.println(Math.ulp(63000000.0));
prints
7.450580596923828E-9
However, you can just use `int`s here, because your values, at about 63 million, are still well below the maximum possible `int` value, which is about 2 billion.
Problem
When I assign from an int to a float I thought float allows more precision, so would not lose or change the value assigned, but what I am seeing is something quite different. What is going on here? ``` for(int i = 63000000; i < 63005515; i++) { int a = i; float f = 0; f=a; System.out.print(java.text.NumberFormat.getInstance().format(a) + " : " ); System.out.println(java.text.NumberFormat.getInstance().format(f)); } ``` some of the output : ... 63,005,504 : 63,005,504 63,005,505 : 63,005,504 63,005,506 : 63,005,504 63,005,507 : 63,005,508 63,005,508 : 63,005,508 Thanks!