[英]how to find index of an array in java
i am trying to use this function but it is giving that index is -1.我正在尝试使用此函数,但它给出的索引为 -1。 so what can i do
那我能做什么
int[] veer = {14,25,14,56,15,36,56,77,18,29,49};
int a = Arrays.asList(veer).indexOf(14);
System.out.println("my array at position 14 is :" + (a));
I assume, you want to learn/test what indexOf
is doing.我假设,您想学习/测试
indexOf
正在做什么。 The problem is, that Arrays.asList(veer)
returns list containing one element - whole array.问题是,
Arrays.asList(veer)
返回包含一个元素的列表 - 整个数组。
When I split the code to multiple lines, it is doing:当我将代码拆分为多行时,它正在执行以下操作:
int[] orig = {14,25,14,56,15,36,56,77,18,29,49};
List<int[]> origList = Arrays.asList(orig);
and that's why, your indexOf is not able to find 14 in a list, because simply said, there is no integer in list.这就是为什么,您的 indexOf 无法在列表中找到 14,因为简单地说,列表中没有整数。
When you use wrapper type, it works as you might expect当您使用包装器类型时,它会按您的预期工作
Integer[] veer = {14,25,14,56,15,36,56,77,18,29,49};
List<Integer> list = Arrays.asList(veer);
int a = list.indexOf(25); // I changed to 25
System.out.println("my array at position 25 is :" + a);
which prints index 1.打印索引 1。
To learn about wrapping/unwrapping array, you can read more here Converting Array of Primitives to Array of Containers in Java but changing int[]
to Integer[]
is simplest in this case (you are not forced by API to accept int[]
).要了解包装/解包数组,您可以在此处阅读更多内容Converting Array of Primitives to Array of Containers in Java但在这种情况下将
int[]
更改为Integer[]
是最简单的(API 不会强迫您接受int[]
) .
In your code在你的代码中
int[] veer = {14,25,14,56,15,36,56,77,18,29,49};
to到
Integer[] veer = {14,25,14,56,15,36,56,77,18,29,49};
because indexOf
expect a object
type since it is primitive it is return as -1(-1 represents not found)因为
indexOf
期望一个object
类型,因为它是原始的,所以它返回为 -1(-1 表示未找到)
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