參考下面提供的代碼,在main()方法中,傳遞給靜態sort()函數的實際參數的類型爲String [](變量a),但是類型的形式參數()函數的類型爲Comparable []。由於這兩種類型似乎不匹配,這怎麼可能?是否有某種隱式對象轉換適用於我不知道的函數調用?任何幫助,將不勝感激。函數調用期間的隱式對象轉換
Merge.java
下面是語法從§2.2歸併強調Merge.java的版本。
/******************************************************************************
* Compilation: javac Merge.java
* Execution: java Merge < input.txt
* Dependencies: StdOut.java StdIn.java
*
*
* Sorts a sequence of strings from standard input using mergesort.
*
* % more tiny.txt
* S O R T E X A M P L E
*
* % java Merge < tiny.txt
* A E E L M O P R S T X [ one string per line ]
*
* % more words3.txt
* bed bug dad yes zoo ... all bad yet
*
* % java Merge < words3.txt
* all bad bed bug dad ... yes yet zoo [ one string per line ]
*
******************************************************************************/
/**
* The {@code Merge} class provides static methods for sorting an
* array using mergesort.
* <p>
* For additional documentation, see <a href="http://algs4.cs.princeton.edu/22mergesort">Section 2.2</a> of
* <i>Algorithms, 4th Edition</i> by Robert Sedgewick and Kevin Wayne.
* For an optimized version, see {@link MergeX}.
*
* @author Robert Sedgewick
* @author Kevin Wayne
*/
public class Merge {
// This class should not be instantiated.
private Merge() { }
// stably merge a[lo .. mid] with a[mid+1 ..hi] using aux[lo .. hi]
private static void merge(Comparable[] a, Comparable[] aux, int lo, int mid, int hi) {
// precondition: a[lo .. mid] and a[mid+1 .. hi] are sorted subarrays
assert isSorted(a, lo, mid);
assert isSorted(a, mid+1, hi);
// copy to aux[]
for (int k = lo; k <= hi; k++) {
aux[k] = a[k];
}
// merge back to a[]
int i = lo, j = mid+1;
for (int k = lo; k <= hi; k++) {
if (i > mid) a[k] = aux[j++];
else if (j > hi) a[k] = aux[i++];
else if (less(aux[j], aux[i])) a[k] = aux[j++];
else a[k] = aux[i++];
}
// postcondition: a[lo .. hi] is sorted
assert isSorted(a, lo, hi);
}
// mergesort a[lo..hi] using auxiliary array aux[lo..hi]
private static void sort(Comparable[] a, Comparable[] aux, int lo, int hi) {
if (hi <= lo) return;
int mid = lo + (hi - lo)/2;
sort(a, aux, lo, mid);
sort(a, aux, mid + 1, hi);
merge(a, aux, lo, mid, hi);
}
/**
* Rearranges the array in ascending order, using the natural order.
* @param a the array to be sorted
*/
public static void sort(Comparable[] a) {
Comparable[] aux = new Comparable[a.length];
sort(a, aux, 0, a.length-1);
assert isSorted(a);
}
/***************************************************************************
* Helper sorting function.
***************************************************************************/
// is v < w ?
private static boolean less(Comparable v, Comparable w) {
return v.compareTo(w) < 0;
}
/***************************************************************************
* Check if array is sorted - useful for debugging.
***************************************************************************/
private static boolean isSorted(Comparable[] a) {
return isSorted(a, 0, a.length - 1);
}
private static boolean isSorted(Comparable[] a, int lo, int hi) {
for (int i = lo + 1; i <= hi; i++)
if (less(a[i], a[i-1])) return false;
return true;
}
/***************************************************************************
* Index mergesort.
***************************************************************************/
// stably merge a[lo .. mid] with a[mid+1 .. hi] using aux[lo .. hi]
private static void merge(Comparable[] a, int[] index, int[] aux, int lo, int mid, int hi) {
// copy to aux[]
for (int k = lo; k <= hi; k++) {
aux[k] = index[k];
}
// merge back to a[]
int i = lo, j = mid+1;
for (int k = lo; k <= hi; k++) {
if (i > mid) index[k] = aux[j++];
else if (j > hi) index[k] = aux[i++];
else if (less(a[aux[j]], a[aux[i]])) index[k] = aux[j++];
else index[k] = aux[i++];
}
}
/**
* Returns a permutation that gives the elements in the array in ascending order.
* @param a the array
* @return a permutation {@code p[]} such that {@code a[p[0]]}, {@code a[p[1]]},
* ..., {@code a[p[N-1]]} are in ascending order
*/
public static int[] indexSort(Comparable[] a) {
int n = a.length;
int[] index = new int[n];
for (int i = 0; i < n; i++)
index[i] = i;
int[] aux = new int[n];
sort(a, index, aux, 0, n-1);
return index;
}
{
if (hi <= lo) return;
int mid = lo + (hi - lo)/2;
sort(a, index, aux, lo, mid);
sort(a, index, aux, mid + 1, hi);
merge(a, index, aux, lo, mid, hi);
}
// print array to standard output
private static void show(Comparable[] a) {
for (int i = 0; i < a.length; i++) {
StdOut.println(a[i]);
}
}
/**
* Reads in a sequence of strings from standard input; mergesorts them;
* and prints them to standard output in ascending order.
*
* @param args the command-line arguments
*/
public static void main(String[] args) {
String[] a = StdIn.readAllStrings();
Merge.sort(a);
show(a);
}
}
不需要轉換,'String'實現'Comparable',所以它已經是必需的類型。 – Berger
爲了將來的參考,代碼示例應該是最小的,即只包含與您的問題相關的代碼。因此,如果您對兩行代碼有疑問,請顯示這兩行代碼,而不是整個程序的來源。如果你在代碼之前沒有那麼整齊地總結你的問題,我不會費心去試圖理解這個問題。 – meriton
另外,在stackoverflow上,您可以通過在編輯器中選擇它並按下相應的工具欄按鈕來格式化代碼。這次我爲你做了這個。 – meriton