简体   繁体   English

这个java post增量运算符在做什么?

[英]What is this java post increment operator doing?

So there's a piece of an radix sort implemented Java code that reads as below: 因此,有一个基数排序实现的Java代码,内容如下:

aux[count[a[i]]++] = a[i];

Why use a post increment operator? 为什么要使用后递增运算符? Why not aux[count[a[i]]+1] ? 为什么不选择aux [count [a [i]] + 1] Is the post increment simply to increment the value in count[a[i]] by 1 and store it there? 职位增值仅仅是将count [a [i]]中的值加1并将其存储在那里吗?

radixSort.java radixSort.java

int N = a.length;
int[] count = new int[R+1];
for (int i = 0; i < N; i++)
  count[a[i]+1]++;
for (int r = 0; r < R; r++)
  count[r+1] += count[r];

for (int i = 0; i < N; i++)
  aux[count[a[i]]++] = a[i];
for (int i = 0; i < N; i++)
  a[i] = aux[i];

Is the post increment simply to increment the value in count[a[i]] by 1 and store it there? 职位增值仅仅是将count [a [i]]中的值加1并将其存储在那里吗?

Yes, exactly. 对,就是这样。 There are two side-effects of the statement: one is a modification to an element of aux , and the other is a modification to an element in count . 该语句有两个副作用:一个是对aux元素的修改,另一个是对count元素的修改。

Personally I'd avoid writing it like that - I'd probably write: 我个人会避免这样写-我可能会写:

// We don't use i other than to index into a, so use an
// enhanced for loop instead
for (int value : a)
{
    aux[count[value]] = value;
    count[value]++;
}

Note that even if the change to count weren't required, aux[count[a[i]]+1] wouldn't do the same thing - because aux[count[a[i]]++] refers to the element in aux with index count[a[i]] , before the increment, because ++ is being used as a post-increment here. 请注意,即使不需要更改countaux[count[a[i]]+1]也不会做同样的事情-因为aux[count[a[i]]++]是指元素在带有索引count[a[i]] aux递增之前 ,因为在这里++被用作后递增。

This aux[count[a[i]]++] = a[i]; 这个aux[count[a[i]]++] = a[i]; means: 手段:

first, take a[i] value, and use it as index for count. 首先,取a[i]值,并将其用作计数的索引。 Then that value use as index in aux array and place the value of a[i] to its place. 然后,该值用作aux数组中的索引,并将a[i]的值放置在其位置。 Then increment the value on position count[a[i]] for 1 . 然后将位置count[a[i]]上的值增加1

Its like this: 就像这样:

aux[count[a[i]]] = a[i];
count[a[i]] = count[a[i]] + 1;

but shorter. 但更短。

You have also this: aux[++count[a[i]]] = a[i]; 您还具有以下内容: aux[++count[a[i]]] = a[i]; . This will take value from count[a[i]] and first increment for 1 and then use it as index in aux array, so now we search it as: 这将从count[a[i]]获取值,并首先递增1 ,然后将其用作aux数组中的索引,所以现在将其搜索为:

count[a[i]] = count[a[i]] + 1;
aux[count[a[i]]] = a[i];

As you can see, aux[count[a[i]]+1] is not like aux[count[a[i]]++] , since it will not store increased value of a[i] for 1 in a[i] , and it will take index a[i] + 1 from aux , but aux[count[a[i]]++] takes index of a[i] . 正如可以看到, aux[count[a[i]]+1]是不喜欢aux[count[a[i]]++]的,因为它不会存储增加的值a[i]在1 a[i] ,它将从aux索引a[i] + 1 ,但是aux[count[a[i]]++]接受a[i]索引。

aux[count[a[i]]+1] is similar to aux[++count[a[i]]] but the difference is you didn't increment the value in count[a[i]] . aux[count[a[i]]+1]aux[++count[a[i]]]类似,不同之处在于您没有增加count[a[i]]

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM