android sorting an array

Question

i'm trying to sort my array but from some reason i got the following error in my compareTo function:

Cannot invoke compareTo(int) on the primitive type int

my code:

private void sortFriendList()
{
    Arrays.sort(friendsList, new Comparator<FacebookUser>() {

        @Override
        public int compare(FacebookUser lhs, FacebookUser rhs) {                
            ChatMessage[] firstChatMessages = app.getDatabaseManager().getChatMessages(lhs.getId());
            ChatMessage[] lastChatMessages = app.getDatabaseManager().getChatMessages(rhs.getId());
            int firstLastMessageTime = (int) firstChatMessages[firstChatMessages.length-1].getTime();
            int lastLastMessageTime = (int) lastChatMessages[firstChatMessages.length-1].getTime();
            return firstLastMessageTime.compareTo(lastLastMessageTime);
        }
    });     
 }

the code supposed to sort an array buy time stored in long type (which i've cast to int)

keiki · Accepted Answer · 2012-08-21 08:41:08Z

3

The compareTomethod only works on objects. An primitive int is no object. You have to manually compare the int values and return either negative, zero or positive value, if this values is less, equal or greater than the other value.

Using the Integer wrapper class for that is just wasting resources.

This morning I'm more willing to provide some code:

long firstLastMessageTime = ...
long lastLastMessageTime = ...
if (firstLastMessageTime < lastLastMessageTime) {
    return -1;
}
if (firstLastMessageTime > lastLastMessageTime) {
    return 1;
}
return 0;

In some cases you could do firstLastMessageTime minus lastLastMessageTime, but that maybe lead to errors if you handle with negative values or if the difference can be bigger than a 32bit int value. So to be 100% sure, just use the above method all the time.

Additional comment:

Is the value really a int value? Usually the time is defined as the milliseconds since 1.1.1970 in Java, which as datatype is long.

edited Aug 21, 2012 at 8:41

answered Aug 20, 2012 at 20:28

keiki

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2 Comments

Asaf Nevo Over a year ago

its a long, but i need to return an int doesn't i ?

keiki Over a year ago

Read the definition of the compareTo method you implement and it should be clear what you have to return. And yes you have to return an int value.

Bhesh Gurung · Accepted Answer · 2012-08-20 20:35:58Z

2

Don't cast long to int to avoid loss of precision. Rather box it to Long

long firstLastMessageTime = //...
long lastLastMessageTime = //...
return Long.valueOf(firstLastMessageTime).compareTo(Long.valueOf(lastLastMessageTime));

edited Aug 20, 2012 at 20:35

answered Aug 20, 2012 at 20:28

Bhesh Gurung

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1 Comment

keiki Over a year ago

Isn't it a bit overkill to create two objects instead of two simple ifs?

Vinay Lodha · Accepted Answer · 2012-08-20 20:24:14Z

2

Try something like this

        return Integer.valueof(firstLastMessageTime).compareTo(lastLastMessageTime);

Before invoking comapreTo method you have to wrap int into Integer object.

answered Aug 20, 2012 at 20:24

Vinay Lodha

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1 Comment

user1329572 Over a year ago

A little explanation of what you did may help OP (and others) even more.

greenkode · Accepted Answer · 2012-08-20 20:28:00Z

0

You Cannot invoke compareTo on primitive types, You should use their wrapper classes. Integer rather than int

answered Aug 20, 2012 at 20:28

greenkode

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