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Java通用转换类型不匹配

[英]Java Generic Casting Type Mismatch

public class MaxHeap<T extends Comparable<T>> implements Heap<T>{
private T[] heap;
private int lastIndex;

public void main(String[] args){
    int i;
    T[] arr = {1,3,4,5,2}; //ERROR HERE *******
    foo
}

public T[] Heapsort(T[]anArray, int n){
    // build initial heap
    T[]sortedArray = anArray;
    for (int i = n-1; i< 0; i--){
        //assert: the tree rooted at index is a semiheap
        heapRebuild(anArray, i, n);
        //assert: the tree rooted at index is a heap
    }
    //sort the heap array
    int last = n-1;
    //invariant: Array[0..last] is a heap,
    //Array[last+1..n-1] is sorted
    for (int j=1; j<n-1;j++) {
        sortedArray[0]=sortedArray[last];
        last--;
        heapRebuild(anArray, 0, last);
    }
    return sortedArray;
}

protected void heapRebuild(T[ ] items, int root, int size){
foo
}

}

The error is on the line with " T[arr] = {1,3,4,5,2} " 错误出现在“ T[arr] = {1,3,4,5,2} ”行中

Eclipse complains that there is a: Eclipse抱怨有一个:

"Type mismatch: cannot convert from int to T" “类型不匹配:无法从int转换为T”

I've tried to casting nearly everywhere but to no avail.A simple way out would be to not use generics but instead just ints but that's sadly not an option. 我试图在几乎所有地方进行转换,但都无济于事,一个简单的方法是不使用泛型,而只使用int,但遗憾的是这不是一种选择。 I've got to find a way to resolve the array of ints {1,3,4,5,2} into an array of T so that the rest of my code will work smoothly. 我必须找到一种方法来将整数{1,3,4,5,2}的数组解析为T的数组,以便我的其余代码能够顺利运行。

Declare arr as an Integer[] instead of a T[] . arr声明为Integer[]而不是T[] There are also a couple of other small errors that I fixed here: 我在这里还修复了其他一些小错误:

public static void main(String[] args){
    int i;
    Integer[] arr = {1,3,4,5,2}; //ERROR HERE *******

}

public <T> T[] Heapsort(T[]anArray, int n){
    // build initial heap
    T[]sortedArray = anArray;
    for (int i = n-1; i< 0; i--){
        //assert: the tree rooted at index is a semiheap
        heapRebuild(anArray, i, n);
        //assert: the tree rooted at index is a heap
    }
    //sort the heap array
    int last = n-1;
    //invariant: Array[0..last] is a heap,
    //Array[last+1..n-1] is sorted
    for (int j=1; j<n-1;j++) {
        sortedArray[0]=sortedArray[last];
        last--;
        heapRebuild(anArray, 0, last);
    }
    return sortedArray;
}

protected void heapRebuild(T[ ] items, int root, int size){
   //foo
}

When you use a generic type, you must resolve all the type parameters, ie tell the compiler which concrete types you want to use instead of the placeholder T in your code. 使用泛型类型时,必须解析所有类型参数,即告诉编译器要使用哪些具体类型而不是代码中的占位符T As the others already pointed out, a primitive type like int can't be used as a generic type parameter - it must be a reference type, like Integer . 正如其他人已经指出的那样,像int这样的原始类型不能用作泛型类型参数-它必须是引用类型,例如Integer So you can rewrite your main method into something like 因此,您可以将main方法重写为类似

public static void main(String[] args){
    int i = 5;
    Integer[] arr = {1,3,4,5,2};
    MaxHeap<Integer> maxHeap = new MaxHeap<Integer>();

    maxHeap.heapSort(arr, i);
}

Note that it should be static . 注意它应该是static When you instantiate your class, you have to specify the type parameter Integer as above. 实例化类时,必须如上所述指定类型参数Integer Then you can pass it the array to be sorted. 然后,您可以将要传递的数组传递给它。

A further note: this loop 进一步说明:此循环

for (int i = n-1; i< 0; i--){
    ...
}

will never execute - the loop condition should be i > 0 instead. 永远不会执行-循环条件应改为i > 0

Erm, ever thought about what would happen if you ran this code in a MaxHeap of, say, String objects? 嗯,有没有想过如果在MaxHeap(例如String对象)中运行此代码会发生什么情况?

T does not exist until you actually instantiate the generic class. 在实际实例化泛型类之前,T不存在。 So, it does not make sense to create an array of T with integers if you don't know what T is. 因此,如果您不知道T是什么,那么用整数创建T的数组是没有意义的。

EDIT: Also, generics in Java only work with reference types, and int is a value type. 编辑:另外,Java中的泛型仅适用于引用类型,而int是值类型。 Try using Integer (int's wrapper class) instead. 尝试改为使用Integer(int的包装器类)。

You've already said that T extends Comparable<T> . 您已经说过T extends Comparable<T> "int" does not extend Comparable<T> not matter how you cast it. 无论如何转换,“ int”都不会扩展Comparable<T>

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