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Start your free trialShaun Johnson
1,214 PointsNSArray *_predictions;
The video is unclear as to the definition of the array pointer NSArray *_predictions; and accessing the instance variable _predictions. _predictions is accessed directly, "_predictions", and is not a pointer. So why the pointer notation in the explicit declaration?
3 Answers
Stone Preston
42,016 Points_predictions is of type NSArray, which means it is an object. All objects in objective c are declared as pointers.
Shaun Johnson
1,214 PointsThanks, that makes sense. The code goes on to access "_predictions" (if (_predictions == nil)) directly. Can you lend some insight into how this works, given _predictions is a pointer to an object?
Stone Preston
42,016 Pointsare you asking why you didnt have to dereference the pointer like:
if (*_predictions == nil)
Shaun Johnson
1,214 PointsYes. I NSLog'd _predictions and it is in fact the array I defined. I was expecting the address of the object.
Stone Preston
42,016 Pointsok thats because when using NSLog and passing it a %@ specifier, what that really does is send the description message to that object. the description message (and most other messages) accepts a pointer to an objects not the object itself. So thats why
printf(@"Predictions: %@", _predictions)
doesnt print the memory address of _predictions since the description method accepts a pointer as its argument, not actual objects themselves. Objective c methods do all that stuff for you behind the scenes so you dont usually have to dereference object pointers because of the way the language is designed to work. Its a lot more complicated than that simple explanantion, but you get the idea.