2

我一定错过了一些基本的东西——希望有人能向我指出。

我有两个 NSSet。集合 A 包含 40 个 NSNumber 值，集合 B 包含 20 个 NSNumber 值。集合 A 包含集合 B 的所有 20 个值以及集合不共享的 20 个其他值。

我想做的是：NSSet *newSet = Set A - Set B，其中新集合仅包含集合 A 中未包含在集合 B 中的 20 个值。虽然我使用的是“minusSet”，但我的新集合始终包含所有 40 个原始值。我已经记录了每组 A 和 B 的内容，以确保它们包含我上面描述的内容。

这是我的代码：

1) 创建集合 A（包含 40 个 NSNumber）

NSMutableSet *fetchedThreadIds = [NSMutableSet setWithArray:[threadedJsonMessages valueForKey:@"tid"]];

2) 创建集合 B（包含 20 个 NSNumber，它们也在集合 A 中）

NSSet *savedThreads = threadIdsForSavedThreads([fetchedThreadIds allObjects]);

3) 尝试从集合 A 中减去集合 B：

[fetchedThreadIds minusSet:savedThreads];

4) 将所有剩余对象添加到新数组中：

threadIdsForMessageFetch = [fetchedThreadIds allObjects];

但是，threadIdsForMessageFetch 包含 40 个对象而不是 20 个。当我打印每个集合时，我得到：

A组

{(
2540000000136987400,
2540000000141777055,
2540000000144609986,
2540000000125243232,
2540000000144610000,
2320000003021796109,
2540000000144375484,
2540000000136185765,
2540000000135311778,
2540000000135311816,
2540000000124128112,
2540000000125012438,
2480000001401011569,
2540000000144610015,
2540000000135311744,
2540000000139755359,
2540000000135311848,
2540000000143846308,
2540000000145732667,
2540000000143985955,
2540000000136185740,
2540000000135756434,
2540000000134373148,
2540000000140816249,
2540000000130860748,
2540000000127140093,
2540000000143526120,
2540000000102176392,
2540000000144609962,
2540000000128826163,
2540000000136185694,
2540000000133208691,
2540000000144555891,
2540000000138091838,
2540000000130179358,
2540000000126098342,
2540000000145162237,
2540000000131688927,
2540000000136185715,
2540000000143526101
)}

B组

{(
2540000000136987400,
2540000000138091838,
2540000000143985955,
2540000000144555891,
2540000000140816249,
2540000000145732667,
2540000000144609986,
2540000000144610000,
2540000000144609962,
2540000000144610015,
2540000000141777055,
2540000000143526101,
2540000000143526120,
2480000001401011569,
2540000000136185694,
2540000000145162237,
2540000000143846308,
2320000003021796109,
2540000000144375484,
2540000000139755359
)}

新数组：

<__NSArrayI 0xc4b8e40>(
2540000000136987400,
2540000000141777055,
2540000000144609986,
2540000000125243232,
2540000000144610000,
2320000003021796109,
2540000000144375484,
2540000000136185765,
2540000000135311778,
2540000000135311816,
2540000000124128112,
2540000000125012438,
2480000001401011569,
2540000000144610015,
2540000000135311744,
2540000000139755359,
2540000000135311848,
2540000000143846308,
2540000000145732667,
2540000000143985955,
2540000000136185740,
2540000000135756434,
2540000000134373148,
2540000000140816249,
2540000000130860748,
2540000000127140093,
2540000000143526120,
2540000000102176392,
2540000000144609962,
2540000000128826163,
2540000000136185694,
2540000000133208691,
2540000000144555891,
2540000000138091838,
2540000000130179358,
2540000000126098342,
2540000000145162237,
2540000000131688927,
2540000000136185715,
2540000000143526101
)

减集调用前 A 和 B 的计数如下：

2014-05-14 22:37:00.900 App[35877:60b] fetchedThreadIdsCount = 40
2014-05-14 22:37:00.900 App[35877:60b] savedThreadsCount = 20
4

1 回答 1

5

我试过这个并且它有效，所以我猜还有其他代码丢失或者NSNumbers真的不一样，只是打印的表示是一样的。

NSMutableSet *fetchedThreadIds = [NSMutableSet setWithObjects:@1, @2, @3, @4, nil];
NSSet *savedThreads = [NSSet setWithObjects:@2, @3, nil];
NSLog(@"fetchedThreadIds: %@", fetchedThreadIds);
NSLog(@"savedThreads: %@", savedThreads);

[fetchedThreadIds minusSet:savedThreads];
NSLog(@"fetchedThreadIds: %@", fetchedThreadIds);
NSArray *threadIdsForMessageFetch = [fetchedThreadIds allObjects];
NSLog(@"threadIdsForMessageFetch: %@", threadIdsForMessageFetch);

NSLog 输出：

fetchedThreadIds: {(
3,
2,
1,
4 )}
savedThreads: {(
3,
2 )}
fetchedThreadIds: {(
1,
4 )}
threadIdsForMessageFetch: (
1,
4)

发布为主要用于格式化的答案。

于 2014-05-15T02:38:02.080 回答