Question

我在这里尝试将多个数组添加到一个NSMutableArray中实际上我正在添加具有不同值的相同数组多次到一个NSMutable数组这个代码：

NSMutableArray* wordsArray =[[NSMutableArray alloc] init ];
NSMutableArray* meaningsArray=[[NSMutableArray alloc]init];
NSMutableArray* wordsArrayTemp=[[NSMutableArray alloc]init];
NSMutableArray* meaningsArrayTemp=[[NSMutableArray alloc]init ];
NSMutableArray* allWords =[[NSMutableArray alloc]initWithCapacity:2];
NSMutableArray* allMeanings=[[NSMutableArray alloc]initWithCapacity:2];




 for(int i=0;i<2;i++)
{
    int wordsCounter=0;
     [wordsArrayTemp removeAllObjects];
[meaningsArrayTemp removeAllObjects];


    for(NSString *tmp in wordsArray)
        {  
            NSString *meaning =[meaningsArray objectAtIndex:wordsCounter++];

            subtmp= [ tmp substringWithRange:NSMakeRange(0,1)];

            if([currentTable isEqualToString:@"arabicToEnglish"])
            { 
                if([subtmp isEqualToString:[arabicLetters objectAtIndex:i]])
                {
                    [wordsArrayTemp addObject:tmp];
                    [meaningsArrayTemp addObject:meaning];
                }
            }
            else
                if([subtmp isEqualToString:[englishLetters objectAtIndex:i]])
                { 
                    [wordsArrayTemp addObject:tmp];
                    [meaningsArrayTemp addObject:meaning];
                }    
        }
  [allWords insertObject:wordsArrayTemp atIndex:i];
   // [allWords addObject: wordsArrayTemp];
    [allMeanings addObject:meaningsArrayTemp]; 
  NSLog(@"all words count%i",[[allWords objectAtIndex:i] count]);  
}

问题：这里假设的行为是在allWords数组中有2个不同的值。但实际发生的是，2个值用最后一个索引值填充相同的值。我的意思是[allWords objectAtIndex：0]应该有2000个对象，[allWords objectAtIndex：1]应该有3000个，但是它们都有3000个对象会发生什么!!

我在这里错过了什么？!! 日Thnx

Answer 1

将对象添加到数组时，不会复制对象。你只需保存其内存地址。

基本上，您将相同的临时数组添加到父数组中。并且您对所有阵列进行了相同的操作。

也许这段展开的循环代码会让它更清晰一些。

// create new array on a specific memory address. let's say this address is 0x01
NSMutableArray* wordsArrayTemp=[[NSMutableArray alloc]init];

// first iteration of your loop

// remove all objects from array at memory address 0x01
[wordsArrayTemp removeAllObjects];

// add objects to the array at address 0x01
[wordsArrayTemp addObject:tmp];

// insert array (still at address 0x01) to the parent array 
[allWords insertObject:wordsArrayTemp atIndex:i];
// your allWords array now looks like this: {array@0x01}

// second iteration of your loop

// remove all objects from array at memory address 0x01!!! (still the same array as in the first iteration)
// since it's the same array all objects from the array at [allWords objectAtIndex:0] are removed too
[wordsArrayTemp removeAllObjects];

// add objects to the array at address 0x01
[wordsArrayTemp addObject:tmp];

// insert array (still at address 0x01) to the parent array 
[allWords insertObject:wordsArrayTemp atIndex:i];
// your allWords array now looks like this {array@0x01, array@0x01}

解决方案非常简单。

在for循环的开头，而不是从数组中删除allObjects，创建新的数组只需替换

[wordsArrayTemp removeAllObjects];
[meaningsArrayTemp removeAllObjects];

带

wordsArrayTemp = [NSMutableArray array];
meaningsArrayTemp = [NSMutableArray array];

Answer 2

尝试同时添加一个数组：

[[allWords array] addObject:wordsArray.array];

希望它会有所帮助

Answer 3

试试这个： -

[allWords insertObject:[wordsArrayTemp copy] atIndex:i];

它应该有用。

尝试将多个数组添加到一个NSMutableArray

3 个答案: