分配ivar

Question

为什么我不能简单：

self.list = [[NSArray alloc] initWithObjects: @"Who Hash", @"Bubba Gump Shrimp Étouffée", @"Who Pudding", @"Scooby Snacks", @"Everlasting Gobstopper", @"Green Eggs and Ham", @"Soylent Green", @"Hard Tack", @"Lembas Bread", @"Roast Beast", @"Blancmange", nil];

而不是：

NSArray *array = [[NSArray alloc] initWithObjects:@"Who Hash", @"Bubba Gump Shrimp Étouffée", @"Who Pudding", @"Scooby Snacks", @"Everlasting Gobstopper", @"Green Eggs and Ham", @"Soylent Green", @"Hard Tack", @"Lembas Bread", @"Roast Beast", @"Blancmange", nil];
self.list = array;

Answer 1

你发布的两种方法都在泄漏。

假设list声明为如下所示的属性...

@property (nonatomic, retain) NSArray *list;

然后我们知道，当通过合成的设置器设置列表时，使用self.list = xxx，它将被保留，保留计数为1。此外，通过调用alloc/init，保留计数再次上升到2。话虽如此，我们可以用一种内存安全的方式重写你的两种方法，就像这样……

self.list = [[[NSArray alloc] initWithObjects:@"A", @"B", @"C", nil] autorelease];

或

self.list = [NSArray arrayWithObjects:@"A", @"B", @"C", nil];

或

NSArray *array = [[NSArray alloc] initWithObjects:@"A", @"B", @"C", nil];
self.list = array;
[array release];

Answer 2

实际上你可以的。但是如果list属性被声明为retain，你必须这样做：

self.list = [[[NSArray alloc] initWithObjects: @"Who Hash", @"Bubba Gump Shrimp Étouffée", @"Who Pudding", @"Scooby Snacks", @"Everlasting Gobstopper", @"Green Eggs and Ham", @"Soylent Green", @"Hard Tack", @"Lembas Bread", @"Roast Beast", @"Blancmange", nil] autorelease];

或者甚至更短：

self.list =  [NSArray arrayWithObjects: "Who Hash", @"Bubba Gump Shrimp Étouffée", @"Who Pudding", @"Scooby Snacks", @"Everlasting Gobstopper", @"Green Eggs and Ham", @"Soylent Green", @"Hard Tack", @"Lembas Bread", @"Roast Beast", @"Blancmange", nil];

否则你会被泄露的。