Java Math.random 执行强制转换引起的奇怪问题

Question

今天在测试 Math.random() 函数的时候，发现了一个奇怪的问题，最开始在第一个 for 循环中测试，结果输出全部为 0，以为是 random() 函数伪随机引起的，修改到第二个 for 循环中后发现一切正常。

但在继续测试的过程中（第 3 个 for 循环），发现原来问题是在强制转换时引起，只要不声明某一个 double 类型的值来储存 random() 方法得到的值而直接使用强制转换，一定会得到 0，这是为什么呢？

public class RandomLuckyAudience {
    public static void main(String[] args) {
        for(int i = 0; i < 10; i++){
            int num = (int) Math.random() * 3;
            System.err.println("num = " + num);
        }
//        num = 0
//        num = 0
//        num = 0
//        num = 0
//        num = 0
//        num = 0
//        num = 0
//        num = 0
//        num = 0

        for( int i = 0; i < 10; i++){
            double random =  Math.random();
            double luckyNum = random*3;
            System.out.println("double = " + luckyNum + ", int = " + (int) luckyNum);
        }
//        double = 2.890974551493182, int = 2
//        double = 1.5971466678951056, int = 1
//        double = 1.4348488400741541, int = 1
//        double = 2.4762102978586764, int = 2
//        double = 0.9541699041578021, int = 0
//        double = 2.10040834712865, int = 2
//        double = 2.1036786411003243, int = 2
//        double = 2.78565491928864, int = 2
//        double = 1.380838157492293, int = 1
//        double = 1.52299402873967, int = 1

        for( int i = 0; i < 10; i++){
            double random =  Math.random();
            double luckyNum = (int) random*3;
            System.out.println("double = " + luckyNum + ", int = " + (int) luckyNum);
        }
//        double = 0.0, int = 0
//        double = 0.0, int = 0
//        double = 0.0, int = 0
//        double = 0.0, int = 0
//        double = 0.0, int = 0
//        double = 0.0, int = 0
//        double = 0.0, int = 0
//        double = 0.0, int = 0
//        double = 0.0, int = 0
//        double = 0.0, int = 0
    }
}

Answer 1

运算符优先级的问题，

int num = (int) Math.random() * 3;

相当于

int num = ((int) Math.random()) * 3;

改成下面这个就好了

int num = (int) (Math.random() * 3);