Using enumerateObjectsAtIndexes or a for-loop to iterate through all indexes of an NSArray BEFORE a given index

Question

What's the most concise way to iterate through the indexes of an NSArray that occur before a given index? For example:

NSArray *myArray = @[ @"animal" , @"vegetable" , @"mineral" , @"piano" ];

[myArray enumerateObjectsAtIndexes:@"all before index 2" options:nil 
    usingBlock:^(id obj, NSUInteger idx, BOOL *stop) {
           // this block will be peformed on @"animal" and @"vegetable"
    }];

Also, this should not loop at all if the given index is 0.

What's the most concise, elegant way to do this? So far I've only cobbled together clumsy multi-line answers that use annoying NSRanges and index sets. Is there a better way I'm overlooking?

Any number of ways to implement. Concise does not mean clear. Use what does the actual job well. Don't fight the framework. Iterating NSArray does not require use of that method. That method gives block usage which is great for things blocks are good at. — uchuugaka, Mar 23 '13 at 12:21

score 3 · Answer 1 · answered Mar 23 '13 at 12:05

3

NSArray *myArray = @[ @"animal" , @"vegetable" , @"mineral" , @"piano" ];
NSUInteger stopIndex = 2;

[myArray enumerateObjectsUsingBlock:^(id obj, NSUInteger idx, BOOL *stop) {
    if (idx == stopIndex) {
        *stop = YES; // stop enumeration
    } else {
        // Do something ...
        NSLog(@"%@", obj);
    }
}];

answered Mar 23 '13 at 12:05

Martin R

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This is a good one. Is there a way to do this without the if/else? Something like *stop = (idx == stopIndex) ? (return;) : NO; – zakdances Mar 23 '13 at 12:38
@yourfriendzak: Not really. According to the documentation, you should only assign `YES` to `*stop`. And calling `return` from within a ternary operator is what I would call "obfuscate programming" :-) – Martin R Mar 23 '13 at 12:43
That requirement is due to the concurrent enumeration offered by another variant of the enumeration method. – bbum Mar 23 '13 at 13:36

score 3 · Answer 2 · edited Mar 09 '14 at 17:31

3

[myArray enumerateObjectsAtIndexes:[NSIndexSet indexSetWithIndexesInRange:NSMakeRange(0, idx)]     
                           options:0
                        usingBlock:^(id obj, NSUInteger idx, BOOL *stop) {

}];

edited Mar 09 '14 at 17:31

vikingosegundo

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answered Mar 23 '13 at 12:19

zakdances

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Why `-1` in `NSMakeRange(-1, idx)`? Should't that be `0` ? – Martin R Mar 23 '13 at 12:32
1

@yourfriendzak. also providing nil as argument for (NSEnumerationOption) options: will give you a type cast warning. So rather then giving a nil here, it better that you should provide 0 if you want to ignore the warning. – Arslan Apr 29 '13 at 09:45

score 1 · Accepted Answer · answered Mar 23 '13 at 12:03

1

What about :

index = 2;
for (int i = 0; i < [myArray count] && i < index; ++i) {
   id currObj = [myArray objectAtIndex:i];
   // Do your stuff on currObj;
}

answered Mar 23 '13 at 12:03

giorashc

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Why is `i < [myArray count]` needed? – zakdances Mar 25 '13 at 09:33
if index >= [myArray count] you are in trouble :) – giorashc Mar 25 '13 at 14:57

score 1 · Answer 4 · edited May 23 '17 at 11:57

Personally I'd go with a block-based enumeration as shown by Martin R or yourfriendzak, the accepted answer by giorashc is probably the worst, as it doesn't provide a mutation guard.

I want to add a (correct) fast enumeration example

NSUInteger stopIndex = 2;
NSUInteger currentIndex = 0;
for (MyClass *obj in objArray) {
    if (currentIndex < stopIndex) {
        // do sth...
    } else {
        break;
    }
    ++currentIndex;      
}

Using enumerateObjectsAtIndexes or a for-loop to iterate through all indexes of an NSArray BEFORE a given index

4 Answers4