[英]Fill an Array with generic Lists using supplier in Java 8 throws ClassCastEx b/c of type erasure
我想使用Supplier和Stream.generate將帶有通用列表的數組填充為元素。
看起來像這樣:
Supplier<List<Object>> supplier = () -> new ArrayList<Object>();
List<Object>[] test = (List<Object>[]) Stream.generate(supplier).limit(m).toArray();
錯誤輸出為:
Exception in thread "main" java.lang.ClassCastException: [Ljava.lang.Object; cannot be cast to [Ljava.util.List;
現在,如何使用Java 8提供的技術填充具有泛型類型的數組? 或者這是不可能的(還),我必須以“經典”的方式做到這一點?
此致,Claas M.
編輯
根據@ Water的要求,我使用stream.collect(使用Cast測試數組)和傳統的迭代方法對填充數組/列表進行了一些性能測試。
首先使用列表進行性能測試:
private static int m = 100000;
/**
* Tests which way is faster for LISTS.
* Results:
* 1k Elements: about the same time (~5ms)
* 10k Elements: about the same time (~8ms)
* 100k Elements: new way about 1.5x as fast (~18ms vs ~27ms)
* 1M Elements: new way about 2x as fast (~30ms vs ~60ms)
* NOW THIS IS INTERESTING:
* 10M Elements: new way about .1x as fast (~5000ms vs ~500ms)
* (100M OutOfMemory after ~40Sec)
* @param args
*/
public static void main(String[] args) {
Supplier<String> supplier = () -> new String();
long startTime,endTime;
//The "new" way
startTime = System.currentTimeMillis();
List<String> test1 = Stream.generate(supplier).limit(m ).collect(Collectors.toList());
endTime = System.currentTimeMillis();
System.out.println(endTime - startTime);
//The "old" way
startTime = System.currentTimeMillis();
List<String> test2 = new ArrayList();
Iterator<String> i = Stream.generate(supplier).limit(m).iterator();
while (i.hasNext()) {
test2.add(i.next());
}
endTime = System.currentTimeMillis();
System.out.println(endTime - startTime);
}
第二個使用數組的性能測試:
private static int m = 100000000;
/**
* Tests which way is faster for ARRAYS.
* Results:
* 1k Elements: old way much faster (~1ms vs ~6ms)
* 10k Elements: old way much faster (~2ms vs ~7ms)
* 100k Elements: old way about 2x as fast (~7ms vs ~14ms)
* 1M Elements: old way a bit faster (~50ms vs ~60ms)
* 10M Elements: old way a bit faster (~5s vs ~6s)
* 100M Elements: Aborted after about 5 Minutes of 100% CPU Utilisation on an i7-2600k
* @param args
*/
public static void main(String[] args) {
Supplier<String> supplier = () -> new String();
long startTime,endTime;
//The "new" way
startTime = System.currentTimeMillis();
String[] test1 = (String[]) Stream.generate(supplier).limit(m ).collect(Collectors.toList()).toArray(new String[m]);
endTime = System.currentTimeMillis();
System.out.println(endTime - startTime);
//The "old" way
startTime = System.currentTimeMillis();
String[] test2 = new String[m];
Iterator<String> it = Stream.generate(supplier).iterator();
for(int i = 0; i < m; i++){
test2[i] = it.next();
}
endTime = System.currentTimeMillis();
System.out.println(endTime - startTime);
}
}
正如你所看到的,Water確實是對的 - Cast讓它變慢了。 但對於Lists,新方法更快; 至少從100k-1M元素。 我還是不知道為什么它對於10M Elements這么慢,我真的很想聽到一些評論。
Stream生成器仍然生成你想要的對象,唯一的問題是調用toArray()將返回一個對象數組,並且你不能從Object數組向下轉換為子對象數組(因為你有類似的東西: Object [] {ArrayList,ArrayList})。
這是一個正在發生的事情的例子:
你認為你有這個:
String[] hi = { "hi" };
Object[] test = (Object[]) hi; // It's still a String[]
String[] out = (String[]) test;
System.out.println(out[0]); // Prints 'hi'
但你實際上有:
String[] hi = { "hi" };
Object[] test = new Object[1]; // This is not a String[]
test[0] = hi[0];
String[] out = (String[]) test; // Cannot downcast, throws an exception.
System.out.println(out[0]);
你回到上面的直接塊,這就是你得到一個投射錯誤的原因。
有幾種方法可以解決它。 如果你想查看你的列表,你可以很容易地從它們中制作一個數組。
Supplier<List<Integer>> supplier = () -> {
ArrayList<Integer> a = new ArrayList<Integer>();
a.add(5);
a.add(8);
return a;
};
Iterator<List<Integer>> i = Stream.generate(supplier).limit(3).iterator();
// This shows there are elements you can do stuff with.
while (i.hasNext()) {
List<Integer> list = i.next();
// You could add them to your list here.
System.out.println(list.size() + " elements, [0] = " + list.get(0));
}
如果您正在設置處理該功能,您可以執行以下操作:
Supplier<List<Integer>> supplier = () -> {
ArrayList<Integer> a = new ArrayList<Integer>();
a.add(5);
a.add(8);
return a;
};
Object[] objArr = Stream.generate(supplier).limit(3).toArray();
for (Object o : objArr) {
ArrayList<Integer> arrList = (ArrayList<Integer>) o; // This is not safe to do, compiler can't know this is safe.
System.out.println(arrList.get(0));
}
根據Stream Javadocs你可以使用另一個toArray()方法,如果你想把它變成一個數組,但我還沒有探索過這個函數,所以我不想討論我不知道的事情。
認為問題是你使用toArray()而沒有返回Object []的參數。 看一眼
public <T> T[] toArray(T[] a)
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