类型转换Math.random? [英] Type Casting Math.random?

问题描述

浏览了本网站上的问题，找不到关于将Math.random()方法从double转换为int的类型的答案. 我的问题是，为什么Math.random只返回不带括号的0，而当它包含在括号中时却返回随机数? 代码的第一部分返回0:

Had a look around the questions on this site and could not quite find the answer I was looking for about type casting the Math.random() method from double to int. My question is, why does Math.random only return a 0 without parentheses whereas it returns random numbers when it is contained within the parentheses? The first part of code returns 0:

int number; 
number = (int) Math.random() * 10; 
System.out.println("\nThe random number is " + number);

但是此代码有效:

int number; 
number = (int) (Math.random() * 10); 
System.out.println("\nThe random number is " + number);

应该指出的是，我在类型转换中看到了几段不同的代码，其中一些程序员似乎同时使用了两种转换方式.

It should be noted I have seen a few different pieces of code on typecasting whereby some programmers seem to use both ways of casting.

推荐答案

此代码:

number = (int) Math.random() * 10; 

首先计算:

(int) Math.random()

由于Math.random()返回一个从0到但不包括1的数字，如果将其强制转换为int，则将向下舍入为0.然后将10乘以0便得到0.

Since Math.random() returns a number from 0 up to but not including 1, if you cast it to int, it will round down to 0. Then when you multiply 10 to 0 you get 0.

这篇关于类型转换Math.random?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！