How to initialize NSMutableArray in iOS - ios

Question

In my app there is huge number of array lists. That's why I have added all arrays in one main array list, and I have initialized them using a "for" loop.
I am getting error inside the "for" loop: "Fast enumeration variables can't be modified in ARC by default".
NSMutableArray * MainArray = [[NSMutableArray alloc] initWithObjects:NameArray, IdArray, MasterIdNameArray, MasterIdArray, nil];
for (NSMutableArray * array in MainArray) {
array = [[NSMutableArray alloc] init];
}

Answer 1

Yes, you can not modify the values of array in a fast enumeration, i.e. for(x in Array). The object x becomes constant, hence it would through a warning.
However you can use for(int i=0; i<[MainArray count]; i++) loop to achieve this.
But, wait: Why you want to initialize it after adding it to an array. Do it like this:
//first create all the arrays that you have,
//NameArray
//IdArray
//MasterIdNameArray
//MasterIdNameArray
//then add them in the MainArray
NSMutableArray *nameArray = [NSMutableArray array];
NSMutableArray *idArray = [NSMutableArray array];
NSMutableArray *masterIdNameArray = [NSMutableArray array];
NSMutableArray *masterIdNameArray = [NSMutableArray array];
NSMutableArray *mainArray = [#[nameArray, idArray, masterIdNameArray, masterIdNameArray] mutableCopy];
Note: I renamed all the variable for the shake for Naming Conventions in Objective-C.

Answer 2

SeanChense is correct. You cannot put an array without initializing it.
NSMutableArray * MainArray = [[NSMutableArray alloc]init];
for (int i = 0;i < YOURCOUNTHERE;i++) {
NSMutableArray * array= [[NSMutableArray alloc]init];
[mainArray addObject:array];
}

Answer 3

If your NameArray is nil, MainArray is nil.
You can do it likes:
NSMutableArray *mainArray = [#[] mutableCopy];
for (int i = 0;i < 3;i++) {
[mainArray addObject:[#[] mutableCopy]];
}

Answer 4

In my app there is huge number of array lists. That's why I have added all arrays in one main array list, and I have initialized them using a "for" loop.
You appear to be misunderstanding how variables and reference types work. Maybe the following will help:
Your line of code:
NSMutableArray *mainArray = [[NSMutableArray alloc] initWithObjects:nameArray, idArray, masterIdNameArray, masterIdArray, nil];
copies the references stored in each of the variables nameArray, idArray etc. and stores those references in a new array.
Somewhere you must declared each of these variables, e.g. something like:
NSMutableArray *nameArray;
This declares a variable nameArray which can hold a reference to a mutable array, that is a reference of type NSMutableArray *. The variable is initialised with the value nil - the "no reference" value.
When your first line of code is executed the value in each variable is passed in the method call, not the variable itself, so the call is effectively:
NSMutableArray *mainArray = [[NSMutableArray alloc] initWithObjects: nil, nil, nil, nil, nil];
and mainArray is set to reference a new mutable array with zero elements - as all references before the first nil in the argument list are used as the initial values in the array.
After mainArray has been setup in this way any operation on it has no effect on the values stored in variables nameArray et al. - there is no connection to those variables. In your loop:
for (NSMutableArray *array in mainArray) {
array = [[NSMutableArray alloc] init];
}
The variable array is a new variable, which is set in turn to each of the values in mainArray. The variable array does not become an alias for each of the variables nameArray et al. in turn - mainArray holds values not variables.
HTH and you now understand why your code could never do what you intended - that is set the values stored in the variables nameArray et al..