在下面的代碼中,調用swapBig(a,some number,somenumber),其中a是數組,它被複制到swapBig()的bleh []中。當bleh []中的值交換時,a中的相應值也會更改。爲什麼會發生這種情況,以及如何編寫代碼以便只修改bleh []而不是原始的a []?非常感謝!爲什麼將數組傳遞給另一個方法會更改原始數組?
public static void swapBig(String bleh[], int to, int from){ //switches data
//Actually performing the swaps
String temp;
temp = bleh[to];
bleh[to] = bleh[from];
bleh[from] = temp;
}
public static void quickSort(String a[], String b[], String c[], String d[],
String e[],String f[], int from, int to){
//performing the quickSort
if (from >= to) return;
int p = (from + to)/2;
int i = from;
int j = to;
while (i <= j){
if (a[i].compareTo(a[p]) <= 0)
i++;
else if (a[j].compareTo(a[p]) >= 0)
j--;
else{
swapBig(a, i, j);
swapBig(b, i, j);
swapBig(c, i, j);
swapBig(d, i, j);
swapBig(e, i, j);
swapBig(f, i, j);
i++;
j--;
}
}
if (p<j){
swapBig(a, p, j);
swapBig(b, p, j);
swapBig(c, p, j);
swapBig(d, p, j);
swapBig(e, p, j);
swapBig(f, p, j);
p = j;
}else if (p>i){
swapBig(a, p, i);
swapBig(b, p, i);
swapBig(c, p, i);
swapBig(d, p, i);
swapBig(e, p, i);
swapBig(f, p, i);
p = i;
}
quickSort(a, b, c, d,e,f, from, p-1);
quickSort(a, b, c, d,e,f, p + 1, to);
}
public static void main (String args [])
{
//Asking for options (what to sort by/search for)
System.out.println("Sort or Search?");
String look = promptFor.nextLine();
if (look.equalsIgnoreCase("Sort")){
System.out.println("Sort by First, Last, Instrument, Instrument Family,
Special Title, or University:");
String toSortBy = promptFor.nextLine();
if (toSortBy.equalsIgnoreCase("First"))
quickSort(fname,lname,inst,instFam,title,uni,0,9);
if (toSortBy.equalsIgnoreCase("Last"))
quickSort(lname,fname,inst,instFam,title,uni,0,9);
if (toSortBy.equalsIgnoreCase("Instrument"))
quickSort(inst,lname,fname,instFam,title,uni,0,9);
if (toSortBy.equalsIgnoreCase("Instrument Family"))
quickSort(instFam,lname,inst,fname,title,uni,0,9);
if (toSortBy.equalsIgnoreCase("Special Title"))
quickSort(title,lname,inst,instFam,uni,fname,0,9);
if (toSortBy.equalsIgnoreCase("University"))
quickSort(uni,lname,inst,instFam,title,fname,0,9);
print();
main(null); }
else if (look.equalsIgnoreCase("Search")) {
System.out.println("Which last name do you wish to search for?");
searchFor(promptFor.nextLine());
}
else
{
System.out.println("Command Not Recognized\n");
main(null);
}
}
}
在Java中,方法參數是引用而不是值,這意味着您將引用傳遞給對象,而不是對象的副本。請注意,基元是按值傳遞的。 – m0skit0 2013-02-20 15:44:44