如何簡潔地處理這種情況。在if
聲明,我無法正常釋放contactDictionary
...Release Quickie
NSNumber *pIDAsNumber;
...
NSMutableDictionary *contactDictionary = [NSMutableDictionary dictionaryWithDictionary:[defaults dictionaryForKey:kContactDictionary]];
if (!contactDictionary) {
contactDictionary = [[NSMutableDictionary alloc] initWithCapacity:1];
}
[contactDictionary setObject:pIDAsNumber forKey:[myClass.personIDAsNumber stringValue]];
[defaults setObject:contactDictionary forKey:kContactDictionary];
是的。相反,你應該使用'[[defaults dictionaryForKey:kContactDictionary] mutableCopy]'。 – 2010-04-22 23:46:52
感謝參數爲零時的行爲信息 - 它不在文檔中! – 2012-03-06 02:28:33