Java:双精度浮点数 [英] Java: double vs float

问题描述

在另一个 Bruce Eckel 练习中，我编写的代码采用一个方法并更改另一个类中的值.这是我的代码:

In another Bruce Eckel exercise, the code I've written takes a method and changes value in another class. Here is my code:

class Big {
  float b;
}

public class PassObject {
  static void f(Letter y) {
    y.c = 'z';
  } //end f()
  static void g(Big z) {
    z.b = 2.2;
  }

  public static void main(String[] args ) {
    Big t = new Big();
    t.b = 5.6;
    System.out.println("1: t.b : " + t.b);
    g(x);
    System.out.println("2: t.b: " + t.b);
  } //end main
}//end class

它抛出一个错误，提示可能会损失精度".

It's throwing an error saying "Possible loss of precision."

PassObject.java:13: possible loss of precision
found: double
required : float   z.b = 2.2
passobject.java:20: possible loss of precision
found : double
required : float   t.b = 5.6

double不能也是float吗?

推荐答案

是的，但是你必须指定它们是浮点数，否则它们被视为双精度数:

Yes, but you have to specify that they are floats, otherwise they are treated as doubles:

z.b = 2.2f

数字末尾的f"使其成为浮点数而不是双精度数.

The 'f' at the end of the number makes it a float instead of a double.

Java 不会自动将双精度数缩小为浮点数.

Java won't automatically narrow a double to a float.

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