I can't understand why this
float f = Integer.MAX_VALUE; System.out.println(Integer.MAX_VALUE); System.out.println((int)f); produces the same lines,
As well as why this does
Float f2 = (float) Integer.MAX_VALUE; System.out.println(Integer.MAX_VALUE); System.out.println(f2.intValue()); I mean, mantissa length for floating point number is 2^23-1. How does it manage to keep max_value of integer, which is 2^31 - 1?
How does it manage to keep max_value of integer, which is 2^31 - 1?
It actually doesn't. The value of f is 2147483648.
However, the narrowing primitive conversion from float to int clamps the value. It gets to this part:
Otherwise, one of the following two cases must be true:
The value must be too small (a negative value of large magnitude or negative infinity), and the result of the first step is the smallest representable value of type int or long.
The value must be too large (a positive value of large magnitude or positive infinity), and the result of the first step is the largest representable value of type int or long.
You can see this easily by making the number even bigger:
float f = Integer.MAX_VALUE; f = f * 1000; System.out.println(Integer.MAX_VALUE); // 2147483647 System.out.println((int)f); // 2147483647 Or by casting to long instead, which obviously doesn't need to be clamped at the same point:
float f = Integer.MAX_VALUE; System.out.println(Integer.MAX_VALUE); // 2147483647 System.out.println((long)f); // 2147483648 
long instead. Will add that to my answer, if that's okay with you.
f is 2147483648 on mine system it shows smth like 2.14748365E9 and the next explaination, I really don't get it.2.14748365E9 - but it really is exactly 2147483648. Does the casting to long not make this clearer?
floatis you can avoid it.doublegives you half a trillion times the accuracy. Some suggest you should use BigDecimal (not my preference however)