java 将任何类型的 ArrayList 传递给方法
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Passing an ArrayList of any type to a method
提问by codingatty
I'm working on my first Android app, a math game for my kid, and am learning Java in the process. I have two ArrayLists, one of integers 0..9and one strings of possible operations, at present, just +and -.
我正在开发我的第一个 Android 应用程序，一个为我孩子准备的数学游戏，并在此过程中学习 Java。我有两个ArrayLists，一个整数0..9和一个可能的操作字符串，目前，只是+和-。
I would like to write a method that returns a random index into an ArrayList, so I can select a random element. What I'm running into is that I need two methods, one for each type of ArrayList, even though the code is identical. Is there a way to do this in a single method?
我想编写一个将随机索引返回到 的方法ArrayList，以便我可以选择一个随机元素。我遇到的是我需要两种方法，一种用于每种类型的ArrayList，即使代码是相同的。有没有办法在单一方法中做到这一点？
What I use now:
我现在用的：
Random randomGenerator = new Random();
. . .
n = randomIndexInt(possibleOperands);
int op1 = possibleOperands.get(n);
n = randomIndexInt(possibleOperands);
int op2 = possibleOperands.get(n);
n = randomIndexStr(possibleOperations);
String operation = possibleOperations.get(n);
. . .
int randomIndexInt(ArrayList<Integer> a){
int n = randomGenerator.nextInt(a.size());
return n;
}
int randomIndexStr(ArrayList<String> a){
int n = randomGenerator.nextInt(a.size());
return n;
}
What I'd like to do is collapse randomIndexIntand randomIndexStrinto a single method.
我想要做的就是崩溃randomIndexInt，并randomIndexStr成一个单一的方法。
回答by didxga
declare your method as int randomIndexInt(ArrayList<?> a)
将您的方法声明为 int randomIndexInt(ArrayList<?> a)
回答by Nermeen
You need only the size of the array right? so do it:
你只需要数组的大小对吗？这样做：
int randomIndex(int size){
int n = randomGenerator.nextInt(size);
return n;
}
回答by neworld
with this code you can pass any type of list
使用此代码，您可以传递任何类型的列表
int randomIndex(List<?> list){
int n = randomGenerator.nextInt(list.size());
return n;
}
回答by Priyank Doshi
More generic is to use Listthan ArrayListin method signature.
List比ArrayList在方法签名中使用更通用。
int your_method(List<?> a){
//your code
}
回答by Anders Metnik
just make it:
只要做到：
private int randomIndex(int size){
return randomGenerator(size);
}
And then call them with randomIndex(yourArray.size());
然后打电话给他们 randomIndex(yourArray.size());
回答by sunil
you can use Genericsas follows
你可以使用Generics如下
declare your function as below
声明你的功能如下
private <T> int randomIndex(ArrayList<T> a){
int n = randomGenerator.nextInt(a.size());
return n;
}
now you can pass ArrayList<String>or ArrayList<Integer>to this function without any issue like this
现在你可以传递ArrayList<String>或传递ArrayList<Integer>到这个函数而没有像这样的任何问题
ArrayList<String> strList = new ArrayList<String>();
strList.add("on1");
System.out.println(randomIndex(strList));
ArrayList<Integer> intList = new ArrayList<Integer>();
intList.add(1);
System.out.println(randomIndex(intList));
回答by Shark
int randomIndex1(ArrayList<?> a)
{
int n = randomGenerator.nextInt(a.size());
return n;
}
int randomIndex2(int size)
{
int n = randomGenerator.nextInt(size);
return n;
}
