使用 Java 索引数组中的元素

Index of an element in an array using Java

提问by Somdutta

The following code in Java return -1. I thought that it should return 3.

Java 中的以下代码返回 -1。我认为它应该返回 3。

int[] array = {1,2,3,4,5,6}; 
System.out.println(Arrays.asList(array).indexOf(4));

Can you help me understand how this function works.

你能帮我理解这个功能是如何工作的吗？

Thanks

谢谢

回答by Paul Boddington

Before Java 5, Arrays.asListused to accept an Object[]. When generics and varargs were introduced into the language this was changed to

在 Java 5 之前，Arrays.asList用于接受Object[]. 当泛型和可变参数被引入语言时，这被更改为

public static <T> List<T> asList(T... arr)

In your example, Tcannot be intbecause intis a primitive type. Unfortunately, the signature matches with Tequal to int[]instead, which is a reference type. The result is that you end up with a Listcontaining an array, not a Listof integers.

在您的示例中，T不能是int因为int是原始类型。不幸的是，签名与T等于int[]相反，这是一个引用类型。结果是你最终得到一个List包含一个数组，而不是一个List整数。

In Java 4, your code would not have compiled because an int[]is not an Object[]. Not compiling is preferable to producing a strange result, and in Effective Java, Josh Bloch says retrofitting asListto be a varargs method was a mistake.

在 Java 4 中，您的代码不会被编译，因为 anint[]不是Object[]. 不编译比产生奇怪的结果更可取，在 Effective Java 中，Josh Bloch 说改型asList为可变参数方法是一个错误。

回答by Jean-Fran?ois Savard

Arrays.asListexpect an object (or more). Your array (array) is a sole object, thus calling the method will create a list of only one object, which is your array.

Arrays.asList期待一个对象（或更多）。您的数组 ( array) 是一个唯一对象，因此调用该方法将创建一个只有一个对象的列表，这就是您的数组。

Calling indexOfon your list will return -1 because 4will never be found as your list contains the array object, not the list of datas that your array contains.

调用indexOf您的列表将返回 -1，因为4永远不会被找到，因为您的列表包含数组对象，而不是数组包含的数据列表。

回答by glglgl

int[] array = {1,2,3,4,5,6};
Arrays.stream(array).boxed().collect(Collectors.toList()).indexOf(4);

should do what you want while keeping the original int[].

应该做你想做的，同时保持原来的int[]。

You get an IntStreamof it, box it into a Stream<Integer>, collect it to a List<Integer>and then can do an .indexOf()search.

你得到一个IntStream，把它装进一个Stream<Integer>，把它收集到一个List<Integer>，然后可以进行.indexOf()搜索。

回答by Shar1er80

You've got answers as as to what is happening to your array when you use Arrays.asList().

当您使用Arrays.asList().

Other things to consider:

其他需要考虑的事项：

• Since your array is sorted, consider Arrays.binarySearch()
• If your array is not sorted, a simple custom indexOf method can perform the same without wasting cycles converting the int[]to a List<Integer>

• 由于您的数组已排序，请考虑 Arrays.binarySearch()
• 如果阵列未排序，一个简单的自定义的indexOf方法可以在不浪费周期转换执行相同int[]的List<Integer>

Example:

例子：

public static void main(String[] args) throws Exception {
    int[] array = {1, 2, 3, 4, 5, 6};
    System.out.print("Binary Search: ");
    System.out.println(Arrays.binarySearch(array, 4));

    int[] array2 = {5, 8, 2, 5, 3, 4, 1};
    System.out.print("Manual Search: ");
    System.out.println(indexOf(array2, 4));
}

public static int indexOf(int[] array, int search) {
    for (int i = 0; i < array.length; i++) {
        if (array[i] == search) {
            return i;
        }
    }
    return -1;
}

Results:

结果：

Binary Search: 3
Manual Search: 5

回答by Oghli

You cannot use primitive types such as int as parameters to generic classes in Java since int is not a class,you shoud use Integer instead :

您不能使用原始类型（例如 int）作为 Java 中泛型类的参数，因为 int 不是类，您应该使用 Integer 代替：

Integer[] array = {1,2,3,3,4,5}; 

List Arraylist=Arrays.asList(array);
System.out.println(Arraylist.indexOf(1));
System.out.println(Arraylist.indexOf(4));

output:

输出：

0
4

note that Arraylist.indexOf(i)return the index of first occurrence of the element iin list :

请注意，Arraylist.indexOf(i)返回元素i在列表中第一次出现的索引：

System.out.println(Arraylist.indexOf(3));

it will return 2which is first occurrence of element 3 , not 3since there is 2 elements of 3 in list.

它将返回2第一个出现的元素 3 ，而不是3因为列表中有 2 个元素 3 。

回答by Ralph

Since you are using an array of primitive type, this will not work. The array is of type int. Were you using Integertype for example, that will work. -1 means the index of the specified element was not found.

由于您使用的是原始类型的数组，因此这将不起作用。该数组的类型为 int。Integer例如，您是否使用类型，这将起作用。-1 表示未找到指定元素的索引。

This is what you want to do:

这是你想要做的：

Integer[] array = {1,2,3,4,5,6};
System.out.println(Arrays.asList(array).indexOf(4));

回答by Sean

Because when you do this with a primary data type array, it will be treated as the first element in the List.

因为当您使用主要数据类型数组执行此操作时，它将被视为 List 中的第一个元素。

int[] array = {1,2,3,4,5,6};
System.out.println(Arrays.asList(array));

You will get this:

你会得到这个：

[[I@474e8d67]

回答by CoderCroc

int[] array = {1,2,3,4,5,6};

• In java arrayis considered as an Object
• int[]is particularly Objectand not Object[]
• To call Arrays.asListyou need zero or more Objects

• 在java中array被认为是Object
• int[]是特别Object而不是Object[]
• 打电话给Arrays.asList你需要零个或多个Objects

Your method gets called in this way,

你的方法以这种方式被调用，

Arrays.asList((Object)(array));//Yes
Arrays.asList(1,2,3,4,5,6);//No

So you will have List<int[]>and not List<Integer>

所以你会有List<int[]>和没有List<Integer>

Now, you might have asked your self that if Arrays.asList(1,2,3,4)works than why not Arrays.asList(arrayInt)?

现在，您可能会问自己，如果Arrays.asList(1,2,3,4)有效，为什么不有效Arrays.asList(arrayInt)？

In Arrays.asList(1,2,3,4)all the values are autoboxedin to the Integerwhich is solely an Objectso in above case Integer[]cad do the trick.

在Arrays.asList(1,2,3,4)所有的值都被自动装箱到，Integer这只是Object在上面的情况下Integer[]cad 做的伎俩。