need to use IOServiceAddMatchingNotification routine for supporing multiple product identifications.
To show the concept, I got part of this code from a site and revised it.Kept it short.
// Set up matching dictionary.
NSMutableDictionary* matchingDictionary;
for (int n = 0; n < numberOfDevices; n++)
{
matchingDictionary = (NSMutableDictionary*)IOServiceMatching(kIOUSBDeviceClassName);
[matchingDictionary setObject:[NSNumber numberWithLong:myVid[n]] forKey:[NSString stringWithUTF8String:kUSBVendorID]];
[matchingDictionary setObject:[NSNumber numberWithLong:myPid[n]] forKey:[NSString stringWithUTF8String:kUSBProductID]];
// Set up a notification callback for device addition on first match.
IOServiceAddMatchingNotification(g_notificationPort, kIOFirstMatchNotification, (CFMutableDictionaryRef)matchingDictionary, deviceAddedCallback, (void*)self, &g_iteratorAdded);
}
I am not sure if it really is correct?. I did not see complains from the xcode and it works.
This requires a nuanced answer - there are three things to note here:
In principle, yes, you need to create distinct matching notifications for each independent match dictionary.
However, it looks like you're expecting only one io_iterator_t to be created and updated with each matching dictionary, as you only have a single variable to store it, g_iteratorAdded. This is not the case. The code shown suffers from a resource leak. Each successful call to IOServiceAddMatchingNotification will create a new iterator, so you will need to retain all of them in an array or so. And then, when you no longer need the notifications (at the latest, when self is dealloc'd, or you'll get callbacks on a freed object!), you need to release all of the iterators.
For matching multiple different USB product IDs but identical vendor IDs, you actually don't need to create multiple match dictionaries and notifications. Instead of kUSBProductID with a single NSNumber/CFNumber, provide a kUSBProductIdsArrayName (aka kUSBHostMatchingPropertyProductIDArray) and specify an array of numbers. (NSArray/CFArray containing a NSNumber/CFNumber for every product ID.)Alternatively, if your product IDs match some hex pattern, you can also use kUSBProductIDMask in conjunction with kUSBProductID: in this case, candidate devices' product IDs will be bitwise masked (&) with the number provided for kUSBProductIDMask before comparing to the kUSBProductID.
If you need to match multiple vendor IDs, you will still need to create a matching notification for each vendor ID, and provide the list of product IDs in the kUSBProductIdsArrayName value for each.
Update: Sample code for array PID match dictionaries
Some rough untested code for dealing with kUSBProductIdsArrayName, assuming your VIDs/PIDs are laid out like this:
static const uint16_t myVid[] = { 0x1234, 0x5555 };
static const size_t numberOfVids = sizeof(myVid) / sizeof(myVid[0]);
static const uint16_t myPid[] = {
// for VID 0x1234
0x1, 0x2, 0x3, 0x1001, 0x1002,
// for VID 0x555
0x100, 0x101,
};
static const unsigned pidsForVid[] = { 5, 2 };
Setting up the matching dictionaries would then look something like this:
unsigned next_pid_index = 0;
for (int n = 0; n < numberOfVids; n++)
{
NSMutableDictionary* matchingDictionary =
(__bridge_transfer NSMutableDictionary*)IOServiceMatching(kIOUSBDeviceClassName);
[matchingDictionary setObject:#(myVid[n]) forKey:#kUSBVendorID];
NSMutableArray* pid_array = [NSMutableArray arrayWithCapacity:pidsForVid[n]];
for (unsigned i = 0; i < pidsForVid[n]; ++i)
{
[pid_array addObject:#(myPid[next_pid_index])];
++next_pid_index;
}
[matchingDictionary setObject:pid_array forKey:#kUSBProductIdsArrayName];
// Set up a notification callback for device addition on first match.
IOReturn result = IOServiceAddMatchingNotification(
g_notificationPort,
kIOFirstMatchNotification,
(__bridge_retained CFMutableDictionaryRef)matchingDictionary,
deviceAddedCallback,
(__bridge void*)self,
&g_iteratorAdded[n]);
assert(result == kIOReturnSuccess);
}
What does ArrayIndexOutOfBoundsException mean and how do I get rid of it?
Here is a code sample that triggers the exception:
String[] names = { "tom", "bob", "harry" };
for (int i = 0; i <= names.length; i++) {
System.out.println(names[i]);
}
Your first port of call should be the documentation which explains it reasonably clearly:
Thrown to indicate that an array has been accessed with an illegal index. The index is either negative or greater than or equal to the size of the array.
So for example:
int[] array = new int[5];
int boom = array[10]; // Throws the exception
As for how to avoid it... um, don't do that. Be careful with your array indexes.
One problem people sometimes run into is thinking that arrays are 1-indexed, e.g.
int[] array = new int[5];
// ... populate the array here ...
for (int index = 1; index <= array.length; index++)
{
System.out.println(array[index]);
}
That will miss out the first element (index 0) and throw an exception when index is 5. The valid indexes here are 0-4 inclusive. The correct, idiomatic for statement here would be:
for (int index = 0; index < array.length; index++)
(That's assuming you need the index, of course. If you can use the enhanced for loop instead, do so.)
if (index < 0 || index >= array.length) {
// Don't use this index. This is out of bounds (borders, limits, whatever).
} else {
// Yes, you can safely use this index. The index is present in the array.
Object element = array[index];
}
See also:
The Java Tutorials - Language Basics - Arrays
Update: as per your code snippet,
for (int i = 0; i<=name.length; i++) {
The index is inclusive the array's length. This is out of bounds. You need to replace <= by <.
for (int i = 0; i < name.length; i++) {
From this excellent article: ArrayIndexOutOfBoundsException in for loop
To put it briefly:
In the last iteration of
for (int i = 0; i <= name.length; i++) {
i will equal name.length which is an illegal index, since array indices are zero-based.
Your code should read
for (int i = 0; i < name.length; i++)
^
It means that you are trying to access an index of an array which is not valid as it is not in between the bounds.
For example this would initialize a primitive integer array with the upper bound 4.
int intArray[] = new int[5];
Programmers count from zero. So this for example would throw an ArrayIndexOutOfBoundsException as the upper bound is 4 and not 5.
intArray[5];
What causes ArrayIndexOutOfBoundsException?
If you think of a variable as a "box" where you can place a value, then an array is a series of boxes placed next to each other, where the number of boxes is a finite and explicit integer.
Creating an array like this:
final int[] myArray = new int[5]
creates a row of 5 boxes, each holding an int. Each of the boxes has an index, a position in the series of boxes. This index starts at 0 and ends at N-1, where N is the size of the array (the number of boxes).
To retrieve one of the values from this series of boxes, you can refer to it through its index, like this:
myArray[3]
Which will give you the value of the 4th box in the series (since the first box has an index of 0).
An ArrayIndexOutOfBoundsException is caused by trying to retrieve a "box" that does not exist, by passing an index that is higher than the index of the last "box", or negative.
With my running example, these code snippets would produce such an exception:
myArray[5] //tries to retrieve the 6th "box" when there is only 5
myArray[-1] //just makes no sense
myArray[1337] //way to high
How to avoid ArrayIndexOutOfBoundsException
In order to prevent ArrayIndexOutOfBoundsException, there are some key points to consider:
Looping
When looping through an array, always make sure that the index you are retrieving is strictly smaller than the length of the array (the number of boxes). For instance:
for (int i = 0; i < myArray.length; i++) {
Notice the <, never mix a = in there..
You might want to be tempted to do something like this:
for (int i = 1; i <= myArray.length; i++) {
final int someint = myArray[i - 1]
Just don't. Stick to the one above (if you need to use the index) and it will save you a lot of pain.
Where possible, use foreach:
for (int value : myArray) {
This way you won't have to think about indexes at all.
When looping, whatever you do, NEVER change the value of the loop iterator (here: i). The only place this should change value is to keep the loop going. Changing it otherwise is just risking an exception, and is in most cases not necessary.
Retrieval/update
When retrieving an arbitrary element of the array, always check that it is a valid index against the length of the array:
public Integer getArrayElement(final int index) {
if (index < 0 || index >= myArray.length) {
return null; //although I would much prefer an actual exception being thrown when this happens.
}
return myArray[index];
}
To avoid an array index out-of-bounds exception, one should use the enhanced-for statement where and when they can.
The primary motivation (and use case) is when you are iterating and you do not require any complicated iteration steps. You would not be able to use an enhanced-for to move backwards in an array or only iterate on every other element.
You're guaranteed not to run out of elements to iterate over when doing this, and your [corrected] example is easily converted over.
The code below:
String[] name = {"tom", "dick", "harry"};
for(int i = 0; i< name.length; i++) {
System.out.print(name[i] + "\n");
}
...is equivalent to this:
String[] name = {"tom", "dick", "harry"};
for(String firstName : name) {
System.out.println(firstName + "\n");
}
In your code you have accessed the elements from index 0 to the length of the string array. name.length gives the number of string objects in your array of string objects i.e. 3, but you can access only up to index 2 name[2],
because the array can be accessed from index 0 to name.length - 1 where you get name.length number of objects.
Even while using a for loop you have started with index zero and you should end with name.length - 1. In an array a[n] you can access form a[0] to a[n-1].
For example:
String[] a={"str1", "str2", "str3" ..., "strn"};
for(int i=0; i<a.length(); i++)
System.out.println(a[i]);
In your case:
String[] name = {"tom", "dick", "harry"};
for(int i = 0; i<=name.length; i++) {
System.out.print(name[i] +'\n');
}
For your given array the length of the array is 3(i.e. name.length = 3). But as it stores element starting from index 0, it has max index 2.
So, instead of 'i**<=name.length' you should write 'i<**name.length' to avoid 'ArrayIndexOutOfBoundsException'.
So much for this simple question, but I just wanted to highlight a new feature in Java which will avoid all confusions around indexing in arrays even for beginners. Java-8 has abstracted the task of iterating for you.
int[] array = new int[5];
//If you need just the items
Arrays.stream(array).forEach(item -> { println(item); });
//If you need the index as well
IntStream.range(0, array.length).forEach(index -> { println(array[index]); })
What's the benefit? Well, one thing is the readability like English. Second, you need not worry about the ArrayIndexOutOfBoundsException
The most common case I've seen for seemingly mysterious ArrayIndexOutOfBoundsExceptions, i.e. apparently not caused by your own array handling code, is the concurrent use of SimpleDateFormat. Particularly in a servlet or controller:
public class MyController {
SimpleDateFormat dateFormat = new SimpleDateFormat("MM/dd/yyyy");
public void handleRequest(ServletRequest req, ServletResponse res) {
Date date = dateFormat.parse(req.getParameter("date"));
}
}
If two threads enter the SimplateDateFormat.parse() method together you will likely see an ArrayIndexOutOfBoundsException. Note the synchronization section of the class javadoc for SimpleDateFormat.
Make sure there is no place in your code that are accessing thread unsafe classes like SimpleDateFormat in a concurrent manner like in a servlet or controller. Check all instance variables of your servlets and controllers for likely suspects.
You are getting ArrayIndexOutOfBoundsException due to i<=name.length part. name.length return the length of the string name, which is 3. Hence when you try to access name[3], it's illegal and throws an exception.
Resolved code:
String[] name = {"tom", "dick", "harry"};
for(int i = 0; i < name.length; i++) { //use < insteadof <=
System.out.print(name[i] +'\n');
}
It's defined in the Java language specification:
The public final field length, which contains the number of components
of the array. length may be positive or zero.
That's how this type of exception looks when thrown in Eclipse. The number in red signifies the index you tried to access. So the code would look like this:
myArray[5]
The error is thrown when you try to access an index which doesn't exist in that array. If an array has a length of 3,
int[] intArray = new int[3];
then the only valid indexes are:
intArray[0]
intArray[1]
intArray[2]
If an array has a length of 1,
int[] intArray = new int[1];
then the only valid index is:
intArray[0]
Any integer equal to the length of the array, or bigger than it: is out of bounds.
Any integer less than 0: is out of bounds;
P.S.: If you look to have a better understanding of arrays and do some practical exercises, there's a video here: tutorial on arrays in Java
For multidimensional arrays, it can be tricky to make sure you access the length property of the right dimension. Take the following code for example:
int [][][] a = new int [2][3][4];
for(int i = 0; i < a.length; i++){
for(int j = 0; j < a[i].length; j++){
for(int k = 0; k < a[j].length; k++){
System.out.print(a[i][j][k]);
}
System.out.println();
}
System.out.println();
}
Each dimension has a different length, so the subtle bug is that the middle and inner loops use the length property of the same dimension (because a[i].length is the same as a[j].length).
Instead, the inner loop should use a[i][j].length (or a[0][0].length, for simplicity).
For any array of length n, elements of the array will have an index from 0 to n-1.
If your program is trying to access any element (or memory) having array index greater than n-1, then Java will throw ArrayIndexOutOfBoundsException
So here are two solutions that we can use in a program
Maintaining count:
for(int count = 0; count < array.length; count++) {
System.out.println(array[count]);
}
Or some other looping statement like
int count = 0;
while(count < array.length) {
System.out.println(array[count]);
count++;
}
A better way go with a for each loop, in this method a programmer has no need to bother about the number of elements in the array.
for(String str : array) {
System.out.println(str);
}
ArrayIndexOutOfBoundsException whenever this exception is coming it mean you are trying to use an index of array which is out of its bounds or in lay man terms you are requesting more than than you have initialised.
To prevent this always make sure that you are not requesting a index which is not present in array i.e. if array length is 10 then your index must range between 0 to 9
ArrayIndexOutOfBounds means you are trying to index a position within an array that is not allocated.
In this case:
String[] name = { "tom", "dick", "harry" };
for (int i = 0; i <= name.length; i++) {
System.out.println(name[i]);
}
name.length is 3 since the array has been defined with 3 String objects.
When accessing the contents of an array, position starts from 0. Since there are 3 items, it would mean name[0]="tom", name[1]="dick" and name[2]="harry
When you loop, since i can be less than or equal to name.length, you are trying to access name[3] which is not available.
To get around this...
In your for loop, you can do i < name.length. This would prevent looping to name[3] and would instead stop at name[2]
for(int i = 0; i<name.length; i++)
Use a for each loop
String[] name = { "tom", "dick", "harry" };
for(String n : name) {
System.out.println(n);
}
Use list.forEach(Consumer action) (requires Java8)
String[] name = { "tom", "dick", "harry" };
Arrays.asList(name).forEach(System.out::println);
Convert array to stream - this is a good option if you want to perform additional 'operations' to your array e.g. filter, transform the text, convert to a map etc (requires Java8)
String[] name = { "tom", "dick", "harry" };
--- Arrays.asList(name).stream().forEach(System.out::println);
--- Stream.of(name).forEach(System.out::println);
ArrayIndexOutOfBoundsException means that you are trying to access an index of the array that does not exist or out of the bound of this array. Array indexes start from 0 and end at length - 1.
In your case
for(int i = 0; i<=name.length; i++) {
System.out.print(name[i] +'\n'); // i goes from 0 to length, Not correct
}
ArrayIndexOutOfBoundsException happens when you are trying to access
the name.length indexed element which does not exist (array index ends at length -1). just replacing <= with < would solve this problem.
for(int i = 0; i < name.length; i++) {
System.out.print(name[i] +'\n'); // i goes from 0 to length - 1, Correct
}
According to your Code :
String[] name = {"tom", "dick", "harry"};
for(int i = 0; i<=name.length; i++) {
System.out.print(name[i] +'\n');
}
If You check
System.out.print(name.length);
you will get 3;
that mean your name length is 3
your loop is running from 0 to 3
which should be running either "0 to 2" or "1 to 3"
Answer
String[] name = {"tom", "dick", "harry"};
for(int i = 0; i<name.length; i++) {
System.out.print(name[i] +'\n');
}
Each item in an array is called an element, and each element is accessed by its numerical index. As shown in the preceding illustration, numbering begins with 0. The 9th element, for example, would therefore be accessed at index 8.
IndexOutOfBoundsException is thrown to indicate that an index of some sort (such as to an array, to a string, or to a vector) is out of range.
Any array X, can be accessed from [0 to (X.length - 1)]
I see all the answers here explaining how to work with arrays and how to avoid the index out of bounds exceptions. I personally avoid arrays at all costs. I use the Collections classes, which avoids all the silliness of having to deal with array indices entirely. The looping constructs work beautifully with collections supporting code that is both easier to write, understand and maintain.
If you use an array's length to control iteration of a for loop, always remember that the index of the first item in an array is 0. So the index of the last element in an array is one less than the array's length.
ArrayIndexOutOfBoundsException name itself explains that If you trying to access the value at the index which is out of the scope of Array size then such kind of exception occur.
In your case, You can just remove equal sign from your for loop.
for(int i = 0; i<name.length; i++)
The better option is to iterate an array:
for(String i : name )
System.out.println(i);
This error is occurs at runs loop overlimit times.Let's consider simple example like this,
class demo{
public static void main(String a[]){
int[] numberArray={4,8,2,3,89,5};
int i;
for(i=0;i<numberArray.length;i++){
System.out.print(numberArray[i+1]+" ");
}
}
At first, I have initialized an array as 'numberArray'. then , some array elements are printed using for loop. When loop is running 'i' time , print the (numberArray[i+1] element..(when i value is 1, numberArray[i+1] element is printed.)..Suppose that, when i=(numberArray.length-2), last element of array is printed..When 'i' value goes to (numberArray.length-1) , no value for printing..In that point , 'ArrayIndexOutOfBoundsException' is occur.I hope to you could get idea.thank you !
You can use Optional in functional style to avoid NullPointerException and ArrayIndexOutOfBoundsException :
String[] array = new String[]{"aaa", null, "ccc"};
for (int i = 0; i < 4; i++) {
String result = Optional.ofNullable(array.length > i ? array[i] : null)
.map(x -> x.toUpperCase()) //some operation here
.orElse("NO_DATA");
System.out.println(result);
}
Output:
AAA
NO_DATA
CCC
NO_DATA
In most of the programming language indexes is start from 0.So you must have to write i<names.length or i<=names.length-1 instead of i<=names.length.
You could not iterate or store more data than the length of your array. In this case you could do like this:
for (int i = 0; i <= name.length - 1; i++) {
// ....
}
Or this:
for (int i = 0; i < name.length; i++) {
// ...
}
This question already has answers here:
Removing from array during enumeration in Swift?
(9 answers)
Closed 6 years ago.
In this loop we have the potential to reduce the number of items in the loop while processing it. This code works fine in Obj-C, but the Swift loops don't get the message that an item has been removed and end up overflowing the array.
In Objective-C, we had:
for(int i = 4; i < staticBlocks.count; i++)
{
PlayerSprite* spr = [staticBlocks objectAtIndex:i];
[spr setPosition:CGPointMake(spr.position.x, spr.position.y-1)];
if(spr.position.y < -1000)
{
[staticBlocks removeObject:spr];
[spr removeFromParent];
}
if(spr.blockTypeIndex == Block_Type_Power_Up)
{
[spr update];
}
}
In Swift I know of these options:
//for i in 4.stride(to: staticBlocks.count, by: 1){ //crashes
//for i in 4..<staticBlocks.count{ //crashes
for var i = 4; i < staticBlocks.count; i += 1 { //works, but is deprecated
let spr = staticBlocks.objectAtIndex(i) as! PlayerSprite
spr.position = CGPointMake(spr.position.x, spr.position.y-1)
if(spr.position.y < -1000)
{
staticBlocks.removeObject(spr)
spr.removeFromParent()
//break
}
if(spr.blockTypeIndex == k.BlockType.PowerUp)
{
spr.update()
}
}
In this specific case, it really isn't a problem for me to use a break statement (which is currently commented out) to kill the loop and prevent the crash, but it doesn't seem like the proper fix. I assume there will come a time when I need to know how do do this correctly. Is there a non deprecated way to do a for loop, one which processes the count each pass?
A related, unanswered question.
Is the for loop condition evalutaed each loop in swift?
I don't know how to link to a specific answer, but this code did what I needed. Marking as duplicate now.
Removing from array during enumeration in Swift?
var a = [1,2,3,4,5,6]
for (i,num) in a.enumerate().reverse() {
a.removeAtIndex(i)
}
This is because in Swift you cannot remove items from an array while you are iterating over it.
From this question Removing from array during enumeration in Swift?
you can see that you should be using the filter function instead of using a for loop.
For example, don't do this:
for (index, aString: String) in enumerate(array) {
//Some of the strings...
array.removeAtIndex(index)
}
Do something like this:
var theStrings = ["foo", "bar", "zxy"]
// Filter only strings that begins with "b"
theStrings = theStrings.filter { $0.hasPrefix("b") }
(Code example from this answer)
Additionally, it should be noted that filter won't update the array, it will return a new one. You can set your array to be equal to that array afterwards.
An additional way to solve the issue, from the same question is to do this:
var a = [1,2,3,4,5,6]
for (i,num) in a.enumerate().reverse() {
a.removeAtIndex(i)
}
print(a)
I need to toggle three different font sizes in the view controller for terms and conditions screen in an endless loop (13 , 15, 17..13, 15, 17..etc..). The screen is pretty simple, just text on full screen and a button in the navigation bar that when pressed, triggers and event handled in action.
The three fonts are represented by three NSString constants.
-(IBAction)toggleFontSize:(id)sender
{
if (self.currentFontIdentifier == regularFontIdentifier)
{
self.currentFontIdentifier = largeFontIdentifier;
}
else if (self.currentFontIdentifier == largeFontIdentifier)
{
self.currentFontIdentifier = smallFontIdentifier;
}
else
{
self.currentFontIdentifier = regularFontIdentifier;
}
self.termsAndConditionsTextView.font = [[BrandingManager sharedManager] fontWithIdentifier:self.currentFontIdentifier];
}
This code works (for now :)), but it's a nice Mediterranean IF yacht.I am wondering if there is some more mature approach. I already see the stakeholders changing their mind and adding a 4th font size. I want it to be manageable better, so basically once they add a new size I would only add it into some Array and that would be it.
Any ideas for a more mature algorithm?
Declare an instance variable for the current selected index and an array for the three fonts (small, regular and large) and try this:
-(IBAction)toggleFontSize:(id)sender {
_currentSelectedIndex = (_currentSelectedIndex + 1) % 3;
self.currentFontIdentifier = _fontIdentifiers[_currentSelectedIndex];
self.termsAndConditionsTextView.font = [[BrandingManager sharedManager] fontWithIdentifier:self.currentFontIdentifier];
}
You may not need currentFontIdentifier property since it can be obtained with _fontIdentifiers[_currentSelectedIndex]
You could use the methods 'indexOfObject and 'lastObject' of the NSArray class, something like:
Using an array of sizes:
NSArray *fontList = #[#"12","14","18"];
Then you could iterate through it using the indexOfObject
NSUInteger ix = [fontList indexOfObject:self.currentFontIdentifier] + 1;
if ([[fontList lastObject] isEqual:self.currentFontIdentifier])
ix=0;
self.currentFontIdentifier = [fontList objectAtIndex:ix];
or
NSUInteger ix = [fontList indexOfObject:self.currentFontIdentifier] + 1;
if (ix >= [fontList count])
ix=0;
self.currentFontIdentifier = [fontList objectAtIndex:ix];
I want to use the basic Objective C stament for (id object in collection) with multiple objects and conditions like this:
for (Origin *origin in [self.fetchedOriginController fetchedObjects] AND Destiny *destiny in [self.fetchedDestinyController fetchedObjects]))
{
NSLog(#"This route starts from %# and ends in %#, origin.name, destiny.name);
}
So the log would be:
This route starts in London and ends in Sidney
This route starts in Madrid and ends in Barcelona
This route starts in Washington and ends in Vienna
(...)
How can this be done?
If you are looking for all combinations just nest:
for (object in collection)
{
for (object2 in collection2)
{
...
}
}
Or are you looking for pairs of objects from two same sized collections? If so create a loop which provides the index:
NSUInteger count = collection.count;
for (NSUInteger ix = 0; ix < count; ix++)
{
id object = collection[ix];
id object2 = collection2[ix];
...
}
If you want to loop over the common pairs of two different sized collections just change the first line to:
NSUInteger count = MIN(collection.count, collection2.count);
If you want something else edit your question to be more explicit.
Something like this would work for what you're trying I think:
for (NSUInteger i = 0; i < collection1.count; i++) {
id object1 = collection1[i];
id object2 = collection2[i];
// continue ...
}
As far as I know, there's no built in syntax to do what you're asking.
This solution makes the assumption that collection1 and collection2 have the same amount of objects, or at least collection2 doesn't exceed collection1 in count. Given your desired syntax, I think you have already planned for this, I just wanted to mention it in case someone else stumbles on this.
What order do you expect this enumeration to happen in? What kind of association between objects from each collection are you looking for?
If you want to take each object from collection and do something with both that object and each object2 in collection2, use nested loops:
for (object in collection) {
for (object2 in collection2) {
// ...
}
}
If you want to go through both collections in order, use two loops.
for (object in collection) { /* ... */ }
for (object2 in collection2) { /* ... */ }
If you expect some sort of mapping between the objects in collection and collection2 (which presumes they have the same count and at least one of them has unique entries), you might look into a data structure that captures that mapping. Then you can iterate across pairs.
NSDictionary *dictionary = [NSDictionary dictionaryWithObjects:collection2 forKeys:collection];
[dictionary enumerateKeysAndObjectsUsingBlock:^(id key, id obj, BOOL *stop){
// ...
}];