Regarding float type precision

Question

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?

Answer 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