为什么 post 递增运算符在此代码中递增 'a' 失败？

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

表达式的计算方式如下（大致上，我没有检查 JLS 的计算顺序，但我认为它是从左到右）：

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。

Answer 2

a++ 表示 "a" 在您编写它并在之后更改它的地方具有旧值。 ++a 表示 "a" 在第一个变化值之后将被计算。

So:
a = 5;
b = 5;
So:
a++ + ++b = (5 (and a + 1 later) + (at first b + 1) 6.

希望你能理解我）我的英语和你的水平差不多Java=)

为什么 post 递增运算符在此代码中递增 'a' 失败？

Why post increment opertor failed to increment 'a' in this code?

java

add

post-increment