java - 为什么在此代码中后递增运算符无法递增 'a'？

Question

package com;

public class Test {

    public static void main(String[] args)
    {
        int a= 11, b = 10;
        a = a++ + ++b;  //why? output is "22 11" and not "23 11"
        System.out.println(a+" "+b);

    }

}

Answer 1

以下是表达式的求值方式(粗略地说，我没有检查 JLS 的求值顺序，但我认为是从左到右):

a = a++ + ++b;  // a is 11, b is 10
a = 11 + ++b;  // a is 12 but its previous value 11 was returned by a++, b is 10
a = 11 + 11;  // a is 12, b is 11 and its updated value was returned by ++b
a = 22;  // a is 12, b is 11 and its updated value was returned by ++b

因此，这是预期的结果，您只需应用这些运算符的定义即可。