关于java：make arrayList.toArray（）返回更具体的类型

make arrayList.toArray() return more specific types

所以，通常情况下，ArrayList.toArray()会返回Object[]的类型……但假设它是对象Custom的Arraylist如何使toArray()返回Custom[]类型而不是Object[]类型？

这样的：

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List<String> list = new ArrayList<String>();

String[] a = list.toArray(new String[0]);

这是写在java6建议：

1	String[] a = list.toArray(new String[list.size()]);

因为我们在内部实现这样大小的数组a realloc无论如何做它为你更好。这是首选的，因为java6空数组，湖.toarray(new MyClass [ 0 ])或.toarray(new MyClass mylist.size)[(])？

如果你的列表是不恰当的A型铸造之前你需要做电话拷贝。这样的：

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List l = new ArrayList<String>();

String[] a = ((List<String>)l).toArray(new String[l.size()]);

相关讨论

它并不真的需要Object[]回报，例如：

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List<Custom> list = new ArrayList<Custom>();
list.add(new Custom(1));
list.add(new Custom(2));

Custom[] customs = new Custom[list.size()];
list.toArray(customs);

for (Custom custom : customs) {
System.out.println(custom);
}

这里是我的Custom类：

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public class Custom {
private int i;

public Custom(int i) {
this.i = i;
}

@Override
public String toString() {
return String.valueOf(i);
}
}

1	arrayList.toArray(new Custom[0]);

download.oracle.com http：/ / / / /文档/ 7 JavaSE util / / / Java的API arraylist.html # % % [ 29 ] 28java.lang.object拷贝

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@SuppressWarnings("unchecked")
public static <E> E[] arrayListToArray(ArrayList<E> list)
{
int s;
if(list == null || (s = list.size())<1)
return null;
E[] temp;
E typeHelper = list.get(0);

try
{
Object o = Array.newInstance(typeHelper.getClass(), s);
temp = (E[]) o;

for (int i = 0; i < list.size(); i++)
Array.set(temp, i, list.get(i));
}
catch (Exception e)
{return null;}

return temp;
}

样本：

1 2	String[] s = arrayListToArray(stringList); Long[] l = arrayListToArray(longList);