java - 关于浮点型精度

Question

我不明白这是为什么

float f = Integer.MAX_VALUE;
System.out.println(Integer.MAX_VALUE);
System.out.println((int)f);

产生相同的行，

以及为什么会这样

Float f2 = (float) Integer.MAX_VALUE;
System.out.println(Integer.MAX_VALUE);
System.out.println(f2.intValue());

我的意思是， float 的尾数长度是 2^23-1。它如何设法保持整数的 max_value，即 2^31 - 1？

Answer 1

How does it manage to keep max_value of integer, which is 2^31 - 1?

其实不然。 f 的值为 2147483648。

然而，narrowing primitive conversion从 float 到 int 限制值。它到达了这一部分:

• 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.

您可以通过将数字更大来轻松地看到这一点:

float f = Integer.MAX_VALUE;
f = f * 1000;
System.out.println(Integer.MAX_VALUE); // 2147483647
System.out.println((int)f); // 2147483647

或者通过转换为 long 来代替，这显然不需要在同一点进行限制:

float f = Integer.MAX_VALUE;
System.out.println(Integer.MAX_VALUE); // 2147483647
System.out.println((long)f); // 2147483648