CODEWITHSUNDEEP
objective-c
timpone
timpone

Reputation: 19979

fast enumeration on a NSArray / NSMutableArray returned from block

I have the following in a success block for an AFNetworking getPath call:

+(void)allItemsWithBlock: (void (^)(NSArray *items)) block
{
   ...
   NSMutableArray *mutableItems = [NSMutableArray array];
   for (NSDictionary *attributes in [responseObject valueForKey:@"data"]) {
      Item *item = [[Item alloc] initWithAttributes:attributes];
      [mutableItems addObject:item];
   } 
   NSLog(@"here is a count: %i", [mutableItems count]);
   if(block){
      block(mutableItems);
   }

and in the block that gets passed in, I have the following but get the error listed as a comment:

[Item allItemsWithBlock:^(NSArray *items){
    for(Item *thisItem in *items){  // The type 'NSArray' is not a pointer to a fast-enumerable object
      NSLog(@"in the block here");
    }
}];

I've read up on trying to fast-enumeration but am not sure what the problem is. Is the NSMutableArray -> NSArray an issue? Is it because this array is created in a block and thus could be seen as possibly still 'open for change'? I have seen code like this before in our projects and doesn't seem to be a problem.

thx for any help

Upvotes: 1

Views: 1451

Answers (1)

Jsdodgers
Jsdodgers

Reputation: 5312

This is because NSArray *items is already a pointer to an array, *items is trying to find a pointer to a pointer, which it is not.

Just replace:

for(Item *thisItem in *items){

with:

for(Item *thisItem in items){

Upvotes: 6

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