NSMutableArray，int值从1到100

Question

This should be dead easy, but somehow it doesn't want to work for me. Using iOS 7 and XCode 5. All I'm trying to do is create an array with values from 1 to 100.

NSMutableArray *array;
for (int i = 0; i < 100; i++)
{
    [array addObject:i];
}

This doesn't work. I get a "Implicit conversion of 'int' to 'id' is disallowed with ARC. I get it, I can't add primitive types to an NSMutableArray.

    [array addObject:@i];

This doesn't work either. I get a "unexpected '@' in program"

    [array addObject:[NSNumber numberWithInt:i]];
    [array addObject:[NSNumber numberWithInteger:i]];

(either case) This "works" (compiles) but it really doesn't "work". The problem with this is that the value from NSNumber is really not a 1-100. What I get for each row is "147212864", 147212832", "147212840"...not what I want.

Lastly:

for (NSNumber *i = 0; i < [NSNumber numberWithInteger:100]; i++)
{
    [array addObject:i];
}

This also doesn't compile. I get an error on the i++. "Arithmetic on pointer to interface 'NSNumber', which is not a constant size for this architecture and platform"

Any suggestions on how to do this extremely simple thing on obj-c?

Answer 1

Either one of these should work:

NSMutableArray *array = [NSMutableArray array];
for (int i = 0; i < 100; i++) {
    [array addObject:@(i)];
}

or

NSMutableArray *array = [NSMutableArray array];
for (int i = 0; i < 100; i++) {
    [array addObject:[NSNumber numberWithInt:i]];
}

Here are the reasons why your code snippets did not work:

1. [array addObject:i] - You cannot add primitives to Cocoa collections
2. [array addObject:@i] - You forgot to enclose the expression i in parentheses
3. NSNumber *i = 0; i < [NSNumber numberWithInteger:100]; i++ NSNumber *i = 0; i < [NSNumber numberWithInteger:100]; i++ - You cannot increment NSNumber without "unwrapping" its value first.

Answer 2

If memory serves, I think you're simply missing parenthesis around the NSNumber shorthand expression.

NSMutableArray *array = [[NSMutableArray alloc] init];
for (NSUInteger i = 0; i < 100; i++)
{
    [array addObject:@(i)];
}

Answer 3

Minimally, @i should be @(i) as described here . You are also forgetting to allocate and initialise your array

NSMutableArray *array = [[NSMutableArray alloc] init];
for (int i = 0; i < 100; i++) {
    [array addObject:@(i)];
}

And since you are getting: "147212864", 147212832", "147212840"...not what I want. , I think you are probably printing out your information wrongly or because the array is unallocated, that's simply garbage. Can you show us how you are outputting?

Answer 4

 NSMutableArray *array = [[NSMutableArray alloc] init];
 NSNumber *myNum;
 for (int i = 0; i < 100; i++) {

     myNum = [[NSNumber alloc]initWithInt:i];
     [array addObject:myNum];
 }
 NSLog(@"%@", array); // 1 - 99 as expected

Worked for me :)

Answer 5

Just saying: Turn on all reasonable warnings in your Xcode project. Then read what the warnings are saying and do something about them. When you write something like

for (NSNumber *i = 0; i < [NSNumber numberWithInteger:100]; i++)

What does a for loop do? An object is in the end a pointer. So you initalise i to nil. Then you compare a pointer with a random pointer: [NSNumber numberWithInteger:100] returns a pointer to an object which could be anywhere in memory, and you compare pointers. Next the i++: No, you can't increment a pointer to an NSNumber. It doesn't make sense.