blocks|key|4347400|text|我不懂Java，但我了解C#。我认为搜索方法会搜索现有的条目。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4347401|不要使用if条件。|offset|length|style|CODE|4347402|在搜索代码和Catch块中捕获异常，执行以下操作：|4347403|System.out.println("The+value+"+%2B+d+%2B+"+was+not+found+and+cannot+be+deleted");|code-block|syntax|javascript|4347404|entityMap^0|0|4|2|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|U|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|V|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

I don't know Java, but I know C#. I think search method searches for existing entries.

Don't use <code>if</code>condition.

Catch the exception in the search code and the in the catch block, do this:

<pre><code>System.out.println("The value " + d + " was not found and cannot be deleted");
</code></pre>

blocks|key|2030854|text|确保变量size等于数组长度nums.length-1;|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2030855|int+size=nums.length-1;|code-block|syntax|javascript|2030856|entityMap^0|4|4|E|E|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Make sure that variable <code>size</code> is equal to the array length <code>nums.length-1;</code>

<pre><code>int size=nums.length-1;
</code></pre>

blocks|key|5816114|text|您的信息不足以完全跟踪代码，但下面这行代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5816115|for(int+n+=+size;+n+>+value;+n--)
++++nums[value]+=+nums[size];|code-block|syntax|javascript|5816116|看起来很可疑。|5816117|根据OP的评论进行编辑：|5816118|while(nums[position]+!=+s){
++++if(position+<+size)
++++++++position%2B%2B;|5816119|好的，简单地把位置增加到大小，这是越界的。(看起来size等于nums.lenght)|offset|length|style|CODE|5816120|BTW我建议你像这样使用ArrayList，因为它为你提供了很多功能：|BOLD|5816121|ArrayList<Integer>+nums+=+new+ArrayList<Integer>();
//....
return+nums.indexOf(s);+//s+is+an+Integer|5816122|entityMap^0|0|0|0|0|0|P|4|V|B|0|0|7|C|9|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|11|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|12|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|14|8|@$O|15|P|16|Q|R]|$O|17|P|18|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|6|7|19|8|@$O|1A|P|1B|Q|U]|$O|1C|P|1D|Q|R]]|9|@]|A|$]]|$1|V|3|W|5|D|7|1E|8|@]|9|@]|A|$E|F]]|$1|X|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]]|Y|$]]

Your information are not enough to trace the code completely but this line:

<pre><code>for(int n = size; n &gt; value; n--)
 nums[value] = nums[size];
</code></pre>

seems very suspicious.

Edit according to OP's comment:

<pre><code>while(nums[position] != s){
 if(position &lt; size)
 position++;
</code></pre>

ok, simply you increase the position up to size, that is out of bound.(Seems <code>size</code> equals <code>nums.lenght</code>)

BTW I suggest you to use <code>ArrayList</code> like this, cause it provides a lot of functionality for you:

<pre><code>ArrayList&lt;Integer&gt; nums = new ArrayList&lt;Integer&gt;();
//....
return nums.indexOf(s); //s is an Integer
</code></pre>

blocks|key|1612477|text|Edit3：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1612478|在这里，我添加了比上次编辑更好的代码。|1612479|这删除了数组元素。通常情况下，我不会为你工作，但如果没有一点技术知识，这是很难解决的代码，至少在我看来，除非你想继续编码和修复。|1612480|您的搜索成功了，但remove实现失败了。如前所述，您的for循环不正确。它不断重复相同的指令(将搜索到的值放在数组的末尾，永远删除之前的值)，结果是你不删除任何重要的东西，甚至丢失信息，如果你想保留它的话。|1612481|下面的代码应该可以帮助你弄清楚如何为你的程序完成预期的功能。|1612482|public+class+SearchTest+{

static+int+nums[]+=+new+int[]+{+1,+2,+3,+4,+5+};

public+static+void+main(String[]+args)+{

++++delete(2);
++++//deletes+the+number+2,+if+found

++++for+(int+i+=+0;+i+<+nums.length;+i%2B%2B)+{

++++++++System.out.println(nums[i]);

++++}

}

public+static+void+delete(int+d)+{
++++int+value+=+search(d);
++++if+(value+==+-1)+{
++++++++System.out.println("The+value+"+%2B+d
++++++++++++++++%2B+"+was+not+found+and+cannot+be+deleted");
++++}+else
++++++++//removes+the+element,+since+it+was+found
++++++++nums+=+removeElement(nums,+value);
}
/**
*http://stackoverflow.com/questions/4870188/delete-item-from-array-and-shrink-array
*+This+is+the+remove+method+from+this+post,+look+at+this+link+for+more
**/

public+static+int[]+removeElement(int[]+original,+int+index)+{
++++int[]+n+=+new+int[original.length+-+1];
++++System.arraycopy(original,+0,+n,+0,+index);
++++System.arraycopy(original,+index+%2B+1,+n,+index,+original.length+-+index
++++++++++++-+1);
++++return+n;
}

private+static+int+search(int+s)+{
++++int+position+=+0;
++++while+(nums[position]+!=+s)+{
//+changed+to+nums.length
++++++++if+(position+<+nums.length)
++++++++++++position%2B%2B;
++++++++else
++++++++++++position+=+-1;
++++}
++++return+position;
}
}|code-block|syntax|javascript|1612483|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|T|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|U|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|V|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Edit3:

Here I have added better code than my last edit.

This has removed the array element.
Normally I wouldn't do work for you but this is hard code to work out without a bit of technical knowledge, at least in my mind, unless you want to keep coding and fixing.

Your search worked, but the remove implementation didn't. As mentioned before, your for loop was not right. It kept repeating the same instruction (placing the searched value at the end of the array, deleting the before value forever) and ended up with you not deleting anything important and even losing information, if you had wanted to keep it.

The code below should help you in figuring out how to complete the intended functions for your program.

<pre><code>public class SearchTest {

static int nums[] = new int[] { 1, 2, 3, 4, 5 };

public static void main(String[] args) {

 delete(2);
 //deletes the number 2, if found

 for (int i = 0; i &lt; nums.length; i++) {

 System.out.println(nums[i]);

 }

}

public static void delete(int d) {
 int value = search(d);
 if (value == -1) {
 System.out.println("The value " + d
 + " was not found and cannot be deleted");
 } else
 //removes the element, since it was found
 nums = removeElement(nums, value);
}
/**
*http://stackoverflow.com/questions/4870188/delete-item-from-array-and-shrink-array
* This is the remove method from this post, look at this link for more
**/

public static int[] removeElement(int[] original, int index) {
 int[] n = new int[original.length - 1];
 System.arraycopy(original, 0, n, 0, index);
 System.arraycopy(original, index + 1, n, index, original.length - index
 - 1);
 return n;
}

private static int search(int s) {
 int position = 0;
 while (nums[position] != s) {
// changed to nums.length
 if (position &lt; nums.length)
 position++;
 else
 position = -1;
 }
 return position;
}
}
</code></pre>

In my program I am first searching for a value, if it exists the array position is given, if not -1 is given but when i try to return the error message in my program that the value entered could not be found therefore not deleted I receive an actual error. This may sound confusing and I can try to explain more if you do not see what I am saying by looking at my code.
Can someone help?

<pre><code>private int search(int s){
 int position = 0;
 while(nums[position] != s){
 if(position &lt; size)
 position++;
 else
 position = -1;
 }
 return position;
}
public void delete(int d){
 int value = search(d);
 if(search(d) == -1){
 System.out.println("The value " + d + " was not found and cannot be deleted");
 }
 else

 for(int n = size; n &gt; value; n--)
 nums[value] = nums[size];
}
</code></pre>

I receive the error

<pre><code>Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: -1
at Numbers.search(Numbers.java:44)
at Numbers.delete(Numbers.java:53)
at Assignment7.main(Assignment7.java:31)
</code></pre>

Search and delete error in array

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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在我的程序中，我首先搜索一个值，如果它存在，给出数组位置，如果没有给出-1，但当我试图在程序中返回错误消息时，输入的值找不到，因此没有删除，我收到了一个实际的错误。这听起来可能令人困惑，如果您看不到我的代码，我可以尝试解释更多。有人能帮帮忙吗？private int search(int s){    int posi...

问搜索和删除数组中的错误
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