CODEWITHSUNDEEP
iphoneiosnsarray
hemant
hemant

Reputation: 1813

Getting Index of an Object from NSArray?

i am trying to get index of an array through indexOfObject method as follows but when i try to log the value to test the index i get a garbage value.. for testing purposes i am having an array with values {57,56,58..} to get an index of lets say 56,

NSNumber *num = [NSNumber numberWithInteger:56];
NSInteger Aindex = [myArray indexOfObject:num];
NSLog(@" %d",Aindex);

the value i get is something like 2323421. what am i possibly doing wrong??

Upvotes: 68

Views: 96273

Answers (7)

Josh O'Connor
Josh O'Connor

Reputation: 4962

If you're using Swift and optionals make sure they are unwrapped. You cannot search the index of objects that are optionals.

Upvotes: 1

Jayasabeen
Jayasabeen

Reputation: 136

indexOfObject methord will get the index of the corresponding string in that array if the string is like @"Test" and you find like @"TEST" Now this will retun an index like a long number

Upvotes: 0

Rajesh Loganathan
Rajesh Loganathan

Reputation: 11247

Try this:

NSArray's indexOfObject: method. Such as the following:

NSUInteger fooIndex = [someArray indexOfObject: someObject];

Upvotes: 5

user866214
user866214

Reputation:

The index returned by indexOfObject will be the first index for an occurence of your object. Equality is tested using isEqual method.
The garbage value you get is probably equal to NSNotFound.
Try testing anIndex against it. The number you are looking for isn't probably in your array : 

NSNumber *num=[NSNumber numberWithInteger:56];
NSInteger anIndex=[myArray indexOfObject:num];
if(NSNotFound == anIndex) {
    NSLog(@"not found");
}

or log the content of the array to be sure :

NSLog(@"%@", myArray);

Upvotes: 145

joe
joe

Reputation: 479

Folks,

When an object is not found in the array the indexOfObject method does NOT return a 'garbage' value. Many systems return an index of -1 if the item is not found.

However, on IOS - because the indexOfObject returns an UNSIGNED int (aka NSUInteger) the returned index must be greater than or equal to zero. Since 'zero' is a valid index there is no way to indicate to the caller that the object was not found -- except by returning an agreed upon constant value that we all can test upon. This constant agreed upon value is called NSNotFound.

The method:

- (NSUInteger)indexOfObject:(id)anObject;

will return NSNotFound if the object was not in the array. NSNotFound is a very large POSITIVE integer (usually 1 minus the maximum int on the platform).

Upvotes: 40

iCoder4777
iCoder4777

Reputation: 1692

NSNumber *num1 = [NSNumber numberWithInt:56];
    NSNumber *num2 = [NSNumber numberWithInt:57];
    NSNumber *num3 = [NSNumber numberWithInt:58];
    NSMutableArray *myArray = [NSMutableArray arrayWithObjects:num1,num2,num3,nil];
    NSNumber *num=[NSNumber numberWithInteger:58];

    NSInteger Aindex=[myArray indexOfObject:num];

    NSLog(@" %d",Aindex);

Its giving the correct output, may be u have done something wrong with storing objects in ur array.

Upvotes: 6

visakh7
visakh7

Reputation: 26400

I just checked. Its working fine for me. Check if your array has the particular number. It will return such garbage values if element is not present.

Upvotes: 0

Related Questions