1

来自 C++，这是我的问题：

我创建了这种类型的对象：

    Size *one = [[Size alloc] initWithX: 3 andY: 1];
    Size *two = [[Size alloc] initWithX: 4 andY: 7];
    // etc...
    Size *thirtythree = [[Size alloc] initWithX: 5 andY: 9];

（每个对象都有一个@property int x;& ..）@property int y;

我存储在一个数组中，如下所示：

NSArray *arrayOfSizes;

arrayOfSizes = [NSArray arrayWithObjects:one,two,three,four,five,six,
                seven,eight,nine,ten,eleven,twelve,thirteen,
                fourteen,fifteen,sixteen,seventeen,eighteen,
                nineteen,twenty,twentyone,twentytwo,
                twentythree,twentyfour,twentyfive,twentysix,
                twentyseven,twentyeight,twentynine,thirty,
                thirtyone,thirtytwo,thirtythree nil];

现在我有一个类型的对象：

Myobject *myObject = [[Myobject alloc] initWithX: 5 andY: 3];

这也有一个@property int x;& @property int y;...

我想将它的值与数组中找到的对象的值进行比较，直到找到一个具有相似值的数组对象。但我不知道如何在 Obj-C 中做到这一点。（在 C++ 中，我会简单地使用vector v;withv.size();和v[x];..etc ......我想......）

这就是我要找的东西.. :)

while( !wholeOfArrayOfSizesChecked && !found)
{
    if ( // x & y of object in array is equal to x & y of myObject )
    {
        found = YES;
    }
    else if( // whole of array checked)
    {
       wholeOfArrayOfSizesChecked = YES;
    }
    else
    { 
      //move on to the next object of the array..
    }
}

提前感谢您的帮助！

4

6 回答 6

2

好吧，您可以在数组上使用快速枚举。像这样的东西：

Myobject *myObject = [[Myobject alloc] initWithX: 5 andY: 3];

for (Size *s in arrayOfSizes)
{
    if (s.x == myObject.x && s.y == myObject.y)
    {
        // Found one
        // Do something useful...
        break;
    }
}
于 2012-08-09T14:25:54.203 回答
1

另一个：

NSUInteger index = [arrayOfSizes indexOfObjectPassingTest:
    ^BOOL(Size *s, NSUInteger idx, BOOL *stop)
    {
        return (s.x == myObject.x) && (s.y == myObject.y);
    }
];

if (index != NSNotFound) {
    id object = [arrayOfSizes objectAtIndex:index];
}
于 2012-08-09T14:35:40.260 回答
0

for-in 循环怎么样？

for (Size *item in array) {
   // compare 'item' to myObject
   if (/* equal condition here */) break;
}
于 2012-08-09T14:23:49.733 回答
0

尝试这样的事情：

for (int i = 0; i < [arrayOfSizes size]; i++)
{
  Size *current = (Size *)[arrayOfSizes objectAtIndex:i];
  if (myObject.x == current.x && myObject.y == current.y)
  {
    // found
    break;
  }
}
于 2012-08-09T14:25:39.093 回答
0 
-(BOOL) isSize:(Size*)size equalToMyObject:(MyObject*)object{
    return (size.x == object.x) && (size.y == object.y);
}

//In some method where you are checking it:
for (Size* size in arrayOfSizes){
        if ([self isSize:size equalToMyObject:myObject]){
            //You found it! They're equal!
            break;
        }
}
于 2012-08-09T14:30:11.723 回答
0

只是为了使用你给定的结构。不过有更聪明的方法:)

wholeOfArrayOfSizesChecked = NO; 
int currObj = 0  
while( !wholeOfArrayOfSizesChecked && !found)
{
    Size *current = (Size *)[arrayOfSizes objectAtIndex:i];
    if (myObject.x == current.x && myObject.y == current.y)
    {
        found = YES;
    }
    else if(currObj == [arrayOfSizes count] -1 )
    {
       wholeOfArrayOfSizesChecked = YES;
    }
    else
    { 
      currObj++;
    }
}
于 2012-08-09T14:31:00.360 回答