Missing return UITableViewCell

Solution 1:

• Declare the cell at the start of the method,
• assign a value to the cell depending on section and row number,
• throw a fatalError() in all cases that "should not occur",
• return the cell.

Also note that the break statements are not needed. The default behavior in Swift is not to fall through to the next case.

override func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> UITableViewCell {

    let cell: SettingsCell

    switch(indexPath.section) {
    case 0:
        switch (indexPath.row) {
        case 0:
            cell = tableView.dequeueReusableCellWithIdentifier("cell0", forIndexPath: indexPath) as! SettingsCell
            cell.backgroundColor = UIColor.redColor()

        case 1:
            cell = tableView.dequeueReusableCellWithIdentifier("cell1", forIndexPath: indexPath) as! SettingsCell
            cell.backgroundColor = UIColor.whiteColor()

        default:
            fatalError("Unexpected row \(indexPath.row) in section \(indexPath.section)")
        }
    case 1:
        switch (indexPath.row) {
        case 0:
            cell = tableView.dequeueReusableCellWithIdentifier("cell10", forIndexPath: indexPath) as! SettingsCell
            cell.backgroundColor = UIColor.redColor()

        case 1:
            cell = tableView.dequeueReusableCellWithIdentifier("cell11", forIndexPath: indexPath) as! SettingsCell
            cell.backgroundColor = UIColor.whiteColor()

        default:
            fatalError("Unexpected row \(indexPath.row) in section \(indexPath.section)")

        }
    default:
        fatalError("Unexpected section \(indexPath.section)")

    }
    return cell
}

The fatalError() error function is marked as @noreturn, so the compiler knows that program execution will not continue from the default cases. (This also helps to find logic errors in the program.)

The compiler verifies that a value is assigned to cell in all other cases.

The possibility to initialize a constant (let cell ...) in this way is new in Swift 1.2.

Alternatively, you can create a cell and return it "immediately" in each case:

override func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> UITableViewCell {

    switch(indexPath.section) {
    case 0:
        switch (indexPath.row) {
        case 0:
            let cell = tableView.dequeueReusableCellWithIdentifier("cell0", forIndexPath: indexPath) as! SettingsCell
            cell.backgroundColor = UIColor.redColor()
            return cell

        case 1:
            let cell = tableView.dequeueReusableCellWithIdentifier("cell1", forIndexPath: indexPath) as! SettingsCell
            cell.backgroundColor = UIColor.whiteColor()
            return cell

        default:
            fatalError("Unexpected row \(indexPath.row) in section \(indexPath.section)")
        }
    case 1:
        switch (indexPath.row) {
        case 0:
            let cell = tableView.dequeueReusableCellWithIdentifier("cell10", forIndexPath: indexPath) as! SettingsCell
            cell.backgroundColor = UIColor.redColor()
            return cell

        case 1:
            let cell = tableView.dequeueReusableCellWithIdentifier("cell11", forIndexPath: indexPath) as! SettingsCell
            cell.backgroundColor = UIColor.whiteColor()
            return cell

        default:
            fatalError("Unexpected row \(indexPath.row) in section \(indexPath.section)")

        }
    default:
        fatalError("Unexpected section \(indexPath.section)")
    }
}

Again, calling fatalError() solves the "missing return expected" compiler error.

This pattern can be useful if there are different kinds of cells (with different classes) created in each case.

Solution 2:

You must return a cell, if the section is number 2 this methods won't be returning anything, well it's the case when u specify secitons more than two. Solution

• Second "if" will be "else" part
• Limit number of section