Displaying an array in xcode

Question

I am trying to display the an array with different factors of a number ("prime"). But instead of giving me the int numbers I always get 0,1,2,3,4,5,... .

        factors.text = @"";

        int factorsNumber;

        NSMutableArray *array;
        array = [NSMutableArray arrayWithCapacity:5];

        for (factorsNumber=1; factorsNumber<=prime; factorsNumber++) {
            if (prime%factorsNumber == 0) {
                [array addObject:[NSString stringWithFormat:@"%d", factorsNumber]];
            }

        }


        for (int i = 0; i < [array count]; i++) {

            [array replaceObjectAtIndex:1 withObject:@"4"];

            NSString *temp = [NSString stringWithFormat:@"%d, ", i, [[array objectAtIndex:i] intValue]];
            factors.text = [factors.text stringByAppendingString:temp];
        }

Weston · Accepted Answer · 2012-07-25 13:12:29Z

1

Replace

NSString *temp = [NSString stringWithFormat:@"%d, ", i, [[array objectAtIndex:i] intValue]];

with

NSString *temp = [NSString stringWithFormat:@"%d, ", [[array objectAtIndex:i] intValue]];

The problem was you were only printing the array index, not the value.

answered Jul 25, 2012 at 13:12

Weston

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2 Comments

Philipp Braun Over a year ago

sorry for the dumb question simply didn't notice!!

Weston Over a year ago

No problem, sometimes it helps to have fresh eyes.

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