iOS TableView is displaying same object in every cell

Question

I have a weird problem, I am setting up a very basic table view to display contacts and I have the tabelviewcontroller as shown below (I tried to include all relevant portions), but for some reason when I run the app I have the same data label appear in each table cell (all seven). Could someone please see the error I can't... thanks a lot

All the following code is from tableviewcontroller.m I make my array in view did load

- (void)viewDidLoad
{
[super viewDidLoad];

// Uncomment the following line to preserve selection between presentations.
// self.clearsSelectionOnViewWillAppear = NO;

// Uncomment the following line to display an Edit button in the navigation bar for this view controller.
// self.navigationItem.rightBarButtonItem = self.editButtonItem;

self.contacts = [NSMutableArray arrayWithCapacity:10];
Contact *contact = [[Contact alloc] init];
contact.name = @"1";
[self.contacts addObject:contact];
contact.name = @"2";
[self.contacts addObject:contact];
contact.name = @"3";
[self.contacts addObject:contact];
contact.name = @"4";
[self.contacts addObject:contact];
contact.name = @"5";
[self.contacts addObject:contact];
contact.name = @"6";
[self.contacts addObject:contact];
contact.name = @"7";
[self.contacts addObject:contact];
}

I have 1 section

- (NSInteger)numberOfSectionsInTableView:(UITableView *)tableView
{
// Return the number of sections.
return 1;
}

I have the number of rows in my array

- (NSInteger)tableView:(UITableView *)tableView numberOfRowsInSection:(NSInteger)section
{
// Return the number of rows in the section.
return [self.contacts count];
}

I create my cells here

- (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath
{
ContactTableViewCell *cell = [tableView dequeueReusableCellWithIdentifier:@"ContactCell"];
Contact *contact = [self.contacts objectAtIndex:[indexPath row]];
cell.contactNameLabel.text = contact.name;
return cell;
}

So I'm not sure why every table cell label says "7" thank you in advance

+1 for your first post being clear and containing relevant info. So rare these days. — rmaddy
– rmaddy, Commented May 23, 2014 at 4:14
@NamoNamo Please don't post a comment asking the OP to look at your answer. The answer itself is enough. — rmaddy
– rmaddy, Commented May 23, 2014 at 4:19

Akhilrajtr · Accepted Answer · 2014-05-23 04:13:02Z

5

In viewDidLoad

- (void)viewDidLoad {

     self.contacts = [NSMutableArray arrayWithCapacity:10];
     Contact *contact = [[Contact alloc] init];
     contact.name = @"1";
     [self.contacts addObject:contact];

     //Create new instance of Contact
     contact = [[Contact alloc] init];
     contact.name = @"2";
     [self.contacts addObject:contact];

     //Add objects same in way
}

If not created new object [[Contact alloc] init], you were changing the same object.

answered May 23, 2014 at 4:13

Akhilrajtr

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1 Comment

shane Over a year ago

wow thanks for the response, I didn't realize adding an object to the array didn't create a copy of it

batman · Accepted Answer · 2014-05-23 04:15:17Z

2

You will need to create a new instance of Contact each time you add it to your data source. Like so

self.contacts = [NSMutableArray arrayWithCapacity:10];
    for(int i=1; i<8;i++) {
        Contact *contact = [[Contact alloc] init];
        contact.name = <some value>;
        [self.contacts addObject:contact];
    }

Hope this helps.

answered May 23, 2014 at 4:15

batman

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Comments

manujmv · Accepted Answer · 2014-05-23 04:15:00Z

1

This is because you are adding same object in each index. You need initialize a new object each time before you adding it to the array. See the below example

Contact *contact1 = [[Contact alloc] init];
contact.name = @"1";
[self.contacts addObject:contact1];
Contact *contact2 = [[Contact alloc] init];
contact.name = @"2";
[self.contacts addObject:contact2];

answered May 23, 2014 at 4:15

manujmv

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Comments

Jitendra · Accepted Answer · 2014-05-23 04:15:55Z

You need to create new object everytime like below code.

self.contacts = [NSMutableArray arrayWithCapacity:10];
Contact *contact = [[Contact alloc] init];
contact.name = @"1";
[self.contacts addObject:contact];

contact = [[Contact alloc] init];
contact.name = @"2";
[self.contacts addObject:contact];

contact = [[Contact alloc] init];

contact.name = @"3";
[self.contacts addObject:contact];

contact = [[Contact alloc] init];

contact.name = @"4";
[self.contacts addObject:contact];

contact = [[Contact alloc] init];

contact.name = @"5";
[self.contacts addObject:contact];

contact = [[Contact alloc] init];

contact.name = @"6";
[self.contacts addObject:contact];

contact = [[Contact alloc] init];

contact.name = @"7";
[self.contacts addObject:contact];

crazyshark · Accepted Answer · 2014-05-23 04:18:10Z

1

You are adding the same object to the array, so basically all the 7 items in your array are just pointers to the same object (with the value 7). Try create a new contact for each number, so like

Contact *contact1 = [[Contact alloc] init];
Contact *contact2 = [[Contact alloc] init];
Contact *contact3 = [[Contact alloc] init];

ect.

answered May 23, 2014 at 4:18

crazyshark

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Comments

Sunny Shah · Accepted Answer · 2014-05-23 04:34:38Z

1

It should be like this

First Contact

 Contact *contact = [[Contact alloc] init];
    contact.name = @"1";
    [self.contacts addObject:contact];

Second Contact

    contact = [[Contact alloc] init];
contact.name = @"2";
[self.contacts addObject:contact];

edited May 23, 2014 at 4:34

answered May 23, 2014 at 4:13

Sunny Shah

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3 Comments

rmaddy Over a year ago

No need for separate variables for each. Just reuse contact.

Sunny Shah Over a year ago

it will understood better for the one who is new to objective c

Sunny Shah Over a year ago

@NamoNamo there are many ways to answer above question. but i prefer the way that he would understood better with this. if u think there is better answer with this then you should give your answer. lots of people says that instead of this you should do this. i appreciate for that but there is no wrong with above answer. if i reuse the code then may be he just copy and paste the answer

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