自动铸造 [英] Automatic casting

问题描述

我必须编写从用户获取一个数字 n 的程序，然后计算和：s = 1/1 + 1/2 + ... + 1 / n。

I have to write program that gets a number n from the user, and then calculates the sum: s = 1/1 + 1/2 + ... + 1/n.

我写了以下代码：

import java.util.Scanner;

public class Test {
    public static void main(String[] args) {
        Scanner unos = new Scanner(System.in);
        System.out.println("n=?");
        int n = unos.nextInt();

        double s = 0.0;
        for (int i = 1; i <= n; i++) {
            s = s + (1.0 / i);
        }
        System.out.println("s=" + s);
    }
}

Java如何决定转换int值 i into double in this statement：

How does Java decide to convert the int value i into double in this statement:

s = s + (1.0 / i);

推荐答案

Java语言规范中定义的其他类型第5章 - 转化和宣传 a>。

The rules that govern what type gets converted/promoted to what other type are defined in the Java Language Spec Chapter 5 - Conversions and Promotions.

对于大多数算术运算，请查看二进制数值推广部分。

Specifically for most arithmetic operations, look at the Binary Numeric Promotion section.

一对操作数，每一个都必须表示数值类型的值，以下规则按顺序使用加宽转换（§5.1.2），以便根据需要转换操作数：

When an operator applies binary numeric promotion to a pair of operands, each of which must denote a value of a numeric type, the following rules apply, in order, using widening conversion (§5.1.2) to convert operands as necessary:

• 如果任一操作数的类型为double，则另一个转换为double。


• 否则，如果任一操作数的类型为float，

在您的情况下，1.0是双精度型，因此 i 被转换为double（加宽转换）。由于 s 已经是双精度，因此不需要进一步的转换。

In your case, 1.0 is a double, so i is converted to a double (widening conversion). Since s already is a double, no further conversion is necessary.

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