数组执行问题困扰我

问题描述

假设我们获得以下代码：



int [] myArray = {4,5,6,1,2,3,21,20} ;

for（int i = 0; i< 5; i ++）{

myArray [i] = myArray [length -1 - i];

}



执行后，myArray数组包含哪些内容？



我尝试了什么：



我不确定这个问题的长度意味着什么。如果长度意味着5，我认为答案是{0，-1，-2,3,2,1，-17，-16}。如果length表示此Array中有多少值，则应为8，答案应为{3,2,1,6,5,4，-14，-13}

Suppose that we are given the following code:

int[] myArray = {4, 5, 6, 1, 2, 3, 21, 20};
for (int i = 0; i < 5; i++) {
myArray[i] = myArray[length -1 - i];
}

After it is executed what will the array myArray contain?

What I have tried:

I am not sure what length means in this question. If length means 5, I think the answer is{0, -1, -2, 3, 2, 1, -17, -16}. If length means how many value in this Array, it should be 8, the answer should be {3, 2, 1, 6, 5, 4, -14, -13}

推荐答案

myArray[length -1 - 0]

然后使用

And then uses

myArray[length -1 - 1]

等等。

所以if长度为5（对myArray来说是错误的 - 它应该是8）然后你加载的索引是4,3,2,1,0

所以元素0变为元素4：2

元素1变为元素3：1

元素2变为元素2：6

元素3变为元素1（其中已经改变了）：1

Elem ent 4成为元素0（已更改）：2



说真的：使用调试器，你会看到我的意思！

And so forth.
So the if length is 5 (which is wrong for myArray - it should be 8) then the indexes you load from are 4, 3, 2, 1, 0
So element 0 becomes element 4: 2
Element 1 becomes element 3: 1
Element 2 becomes element 2: 6
Element 3 becomes element 1 (which has changed): 1
Element 4 becomes element 0 (which has changed): 2

Seriously: use the debugger and you'll see what I mean!

int tmp;
int[] myArray = {4, 5, 6, 1, 2, 3, 21, 20};
for (int i = 0; i < 5; i++) {
    tmp= length -1 - i;
    myArray[i] = myArray[length -1 - i];
}

tmp只会显示下一行获取值的位置。

The tmp will just show you where it get the value in next line.

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