Fast enumeration on an NSIndexSet

Can you fast enumerate a 开发者_如何转开发NSIndexSet? if not, what's the best way to enumerate the items in the set?

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In OS X 10.6+ and iOS SDK 4.0+, you can use the -enumerateIndexesUsingBlock: message:

NSIndexSet *idxSet = ...

[idxSet enumerateIndexesUsingBlock:^(NSUInteger idx, BOOL *stop) {
  //... do something with idx
  // *stop = YES; to stop iteration early
}];

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A while loop should do the trick. It increments the index after you use the previous index.

/*int (as commented, unreliable across different platforms)*/
NSUInteger currentIndex = [someIndexSet firstIndex];
while (currentIndex != NSNotFound)
{
    //use the currentIndex

    //increment
    currentIndex = [someIndexSet indexGreaterThanIndex: currentIndex];
}

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Fast enumeration must yield objects; since an NSIndexSet contains scalar numbers (NSUIntegers), not objects, no, you cannot fast-enumerate an index set.

Hypothetically, it could box them up into NSNumbers, but then it wouldn't be very fast.

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Short answer: no. NSIndexSet does not conform to the <NSFastEnumeration> protocol.

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Supposing you have an NSTableView instance (let's call it *tableView), you can delete multiple selected rows from the datasource (uhm.. *myMutableArrayDataSource), using:

[myMutableArrayDataSource removeObjectsAtIndexes:[tableView selectedRowIndexes]];

[tableView selectedRowIndexes] returns an NSIndexSet. No need to start enumerating over the indexes in the NSIndexSet yourself.

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These answers are no longer true for IndexSet in Swift 5. You can perfectly get something like:

let selectedRows:IndexSet = table.selectedRowIndexes

and then enumerate the indices like this:

for index in selectedRows {
   // your code here.
}