iOS/Swift: How to add 1 to i in a loop for for a specific value and how to retrograd a 'loop for' for specific moment? - Ios swift

lostA
October 6, 2022
218 views
2 votes
3 Answers

I try to add 1 to i in a loop for in a specific moment? When I do this, it comes to the "normal" value instantly at the begining of the loop.

for var i in (0..<10)
    {
    if (i == 2)
       {
       i += 1
       }
    }

EDIT:

Concret code:

                   for var i in (0..<models.count) 
                    {
                        if (i > 0 && models[i].label1 == models[i-1].label1)
                        {
                            models[i-1].slots.append(contentsOf: models[i].slots)
                            models.remove(at: i)
                            i -= 1
                           
                        }
                    }

Answers

- DonMag
- October 6, 2022 at 4:32 pm
- 0 votes
0
The for loop sets the value of i each time through.

So,
```
    for var i in (0..<10)
    {
        // the for loop sets i to the next value in 0 to 9
        print(i)
        if (i == 2)
        {
            // print current value of i
            print("a:", i)
            i += 1
            // print current value of i
            print("b:", i)
        }
    }
```
Inside your if (i == 2) block, you change the value of i but then the for loop changes it right back to the next value in the enumeration of 0 to 9.

The debug output from the above code will look like this:
```
0
1
2
a: 2
b: 3
3
4
5
6
7
8
9
```
To make it more obvious, change i += 1 to i += 4

Output:
```
0
1
2
a: 2
b: 6
3
4
5
6
7
8
9
```
Edit – for a little more clarification…

From the Swift Docs:

For-In Loops

You use the for-in loop to iterate over a sequence, such as items in an array, ranges of numbers, or characters in a string.

Both of these loops:
```
for i in 0..<3 {
    print("i =", i)
}

for i in [0, 1, 2] {
    print("i =", i)
}
```
will output this:
```
i = 0
i = 1
i = 2
```
and this loop:
```
for i in [0, 2, 4] {
    print("i =", i)
}
```
will output this:
```
i = 0
i = 2
i = 4
```
because the for-in loop is iterating over the sequence.

In either case, if we try to modify i inside the loop, we’ll see this in Xcode:

If we change i to a var:
```
for var i in 0..<3 {
    print("a: i =", i)
    i = 7
    print("b: i =", i)
}
```
we can compile and run the code, but… each time through the loop i will be assigned the next value in the iteration, and we’ll get this:
```
a: i = 0
b: i = 7
a: i = 1
b: i = 7
a: i = 2
b: i = 7
```
As Duncan C explained, if we want to actually modify i, the proper thing to do is use a while loop.

So… why can we do this in other languages (such as Objective-C):
```
for (int i = 0; i < 3; i++) {
    NSLog(@"a: i = %d", i);
    i = 7;
    NSLog(@"b: i = %d", i);
}
```
and get this output:
```
a: i = 0
b: i = 7
```
and the loop exits?

That’s because this:

is a more concise yet functionally equivalent way of writing this:

The Swift For-In loop is much more similar to the Objective-C For-In Fast Enumeration loop:
```
for (NSNumber *n in @[@0, @1, @2]) {
    NSLog(@"n = %@", n);
}
```
outputs:
```
n = 0
n = 1
n = 2
```
and, if we try to modify n inside the loop:

so we make it __strong:
```
for (__strong NSNumber *n in @[@0, @1, @2]) {
    NSLog(@"a: n = %@", n);
    n = @7;
    NSLog(@"b: n = %@", n);
}
```
and the output is – as you might guess:
```
a: n = 0
b: n = 7
a: n = 1
b: n = 7
a: n = 2
b: n = 7
```
Hope that helps a bit.
Login or Signup to reply.

- DuncanC
- October 6, 2022 at 4:35 pm
- 0 votes
0
In for loops in Swift, the variable (i) is treated as a let constant, and you cannot mutate it inside the loop.

That seems right. Being able to mutate a loop variable inside the loop can lead to all sorts of unexpected side-effects.

You can use a while loop to get the same effect:

The following works:
```
var i = 0
while i < 10  {
    print(i)
    if i.isMultiple(of: 2) && Int.random(in: 1...3) == 2 {
        i -= 1
    }
    i += 1
}
print("Done")
```
This code throws a compiler error:
```
for index in 1...10 {
    if index.isMultiple(of: 3) {
        index += 1 // <- Error: "left side of mutating operator isn't mutable: 'index' is a 'let' constant"
        print(index)
    }
}
```
Login or Signup to reply.

- JeremyP
- October 6, 2022 at 5:15 pm
- 0 votes
0
The problem is caused because the i is scoped to the block of the loop. Each time round you actually get a new i.

There are three solutions.

The first is to use a while loop as per Duncan C’s answer.

The second is to maintain a separate index and still use a for loop, but that’s really just the same as the first one

The best answer IMO is to count downwards from the end.
```
for i in (1 ..< models.count).reversed()
{
    if (models[i].label1 == models[i-1].label1)
    {
        models[i-1].slots.append(contentsOf: models[i].slots)
        models.remove(at: i)
    }
}
```
Of course, reversing a list of numbers gets expensive for long lists, so you might consider using stride
```
for i in stride(from: models.count - 1, to: 0, by: -1)
{
    if (models[i].label1 == models[i-1].label1)
    {
        models[i-1].slots.append(contentsOf: models[i].slots)
        models.remove(at: i)
    }
}
```
Login or Signup to reply.