增量运算符不会在for循环中递增

本文介绍了增量运算符不会在for循环中递增的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

pre $

 for（int i = 0; i< ; 10; i = i ++）{
 System.err.print（hoo ...）; 
 
 
 
 $ b 这是永无止境的循环！
 
 
任何人都可以解释为什么会发生这种情况？
 int i = 0; i

同样的：
$ b $ pre $ for（int i = 0; i

的第三部分部分>语句 - i = i ++ ，计算方法如下： -

$ $ p $ int oldValue = i; i = i + 1; i = oldValue; //第三步

您需要删除作业从那里开始工作： - $ /
$ b $ pre $ for（int i = 0; i

（来自评论的OP请求）

`x = 1; x = x ++ + x ++;` ： -

就注释中指定的问题而言，以下表达式的结果为： -

  x = 1; 
 x = x ++ + x ++;

获得如下： -

让我们来标记第二条语句的不同部分：

pre $ x $ c $ x = x ++ + x ++;
RAB

现在，首先RHS部分（A + B ）将被评估，然后最终结果将被分配给 x 。所以，让我们继续前进。

第一个 A 会被评估： -

  old1 = x; //`old1变成1` 
 x = x + 1; //增加`x`。 `x变成2` 
 // x = old1; //这不会完成。由于价值尚未分配。

现在，由于将 A 分配给 R 在这里没有完成，第三步是不执行的。

现在，移动到 B 评估： -

  old2 = x; // old2变成2.（因为`x`是2，从`A`的求值）
 x = x + 1; //增加`x`。 `x变成3`。 
 // x = old2; //这将不会在这里完成。

现在，获取 x ++ + x ++ ，我们需要在评估 A 和 B 时做最后一项任务，因为现在是该值被分配在 x 中。为此，我们需要替换： -

  A  - > old1 
 B  - > old2 //两个评估的最后一个任务。 （A和B）
 
 / **参见分解`x = old1;`到最后，理解它如何等于`A = old1;在x = x ++`的情况下，考虑`x ++< ==> A`在这种情况下。 ** /

所以，x = x ++ + x ++

  x = old1 + old2; 
 = 1 + 2; 
 = 3; //因此答案是

 
 
 第三部分 x = x ++ ，看看它是如何工作的 x = x ++ + x ++  case： -  
 
 
 不知道为什么替换完成为 A  - > old1 而不是 x  - > old1 ，例如 x = x ++ 。
 
 
   x = x ++ 部分，特别是最后一项任务： -  $ / 
 
 $ p $  x =属性oldValue; 
  
如果您考虑 x ++ 为 A 在这里，那么上面的赋值可以分解成以下步骤： -  
 
 
  A = oldValue; 
 x = A; 
  
现在，对于目前的问题，它是一样的： -  $ / 
 
 
  A = old1; 
 B = old2; 
 x = A + B; 
  
我希望能够说清楚。
 
I'm doing some research about Java and find this very confusing:
for (int i = 0; i < 10; i = i++) {
  System.err.print("hoo... ");
}
This is never ending loop!
Anybody has good explanation why such thing happens?
 解决方案 
for (int i = 0; i < 10; i = i++) {
The above loop is essentially the same as: -
for (int i = 0; i < 10; i = i) {
the 3 part of your for statement - i = i++, is evaluated as: -
int oldValue = i;
i = i + 1;
i = oldValue;  // 3rd Step
You need to remove the assignment from there, to make it work: -
for (int i = 0; i < 10; i++) {
(On OP request from Comments)
Behaviour of x = 1; x = x++ + x++; : -
As far as your issue as specified in the comment is concerned, the result of the following expression: -
x = 1;
x = x++ + x++;
is obtained as follows: -
Let's mark different parts of the second statement: -
x = x++ + x++;
R    A     B
Now, first the RHS part (A + B) will be evaluated, and then the final result will be assignmed to x. So, let's move ahead.
First A is evaluated: -
old1 = x;  // `old1 becomes 1`
x = x + 1; // Increment `x`. `x becomes 2`
//x = old1; // This will not be done. As the value has not been assigned back yet.
Now, since the assignment of A to R is not done here, the 3rd step is not performed.
Now, move to B evaluation: -
old2 = x;  // old2 becomes 2. (Since `x` is 2, from the evaluation of `A`)
x = x + 1; // increment `x`. `x becomes 3`.
// x = old2; // This will again not be done here.
Now, to get the value of x++ + x++, we need to do the last assignment that we left in the evaluation of A and B, because now is the value being assigned in x. For that, we need to replace: -
A --> old1
B --> old2   // The last assignment of both the evaluation. (A and B)

/** See Break up `x = old1;` towards the end, to understand how it's equivalent to `A = old1; in case of `x = x++`, considering `x++ <==> A` in this case. **/
So, x = x++ + x++, becomes: -
x = old1 + old2;
  = 1 + 2;
  = 3;  // Hence the answer
Break up of 3rd part of x = x++, to see how it works in x = x++ + x++ case: -
Wonder why the replacement is done as A --> old1 and not x --> old1, as in case of x = x++.
Take a deep look at x = x++ part, specially the last assignment: -
x = oldValue;
if you consider x++ to be A here, then the above assignment can be broken into these steps: -
A = oldValue;
x = A;
Now, for the current problem, it is same as: -
A = old1;
B = old2;
x = A + B;
I hope that makes it clear.
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OldValue