Question

我正在尝试制作一款可以检测重复联系人的iPhone应用。我已经做了一个for循环来循环遍历所有联系人（人）的数组，但它似乎只工作一次。

    for (int i = 0; i < [people count]; i++)
    {   
                    ABRecordRef person = (ABRecordRef)[people objectAtIndex:i];
                    ABRecordRef person2;

                    NSString *firstname = (NSString*)ABRecordCopyValue(person, kABPersonFirstNameProperty);
                    NSString *lastname = (NSString*)ABRecordCopyValue(person, kABPersonLastNameProperty);

                    NSLog(@"Firstname = %@", firstname);
                    NSLog(@"People count = %i", [people count]);

                    for (int j = 0; i < [people count]; i++)
                    {
                        person2 = (ABRecordRef)[people objectAtIndex:i];

                        if (j != i 
                            && [firstname isEqualToString:(NSString*)ABRecordCopyValue(person2, kABPersonFirstNameProperty)] 
                            && [lastname isEqualToString:(NSString*)ABRecordCopyValue(person2, kABPersonLastNameProperty)])
                        {
                            //duplicate is found
                            NSString *name = firstname;
                            name = [name stringByAppendingString:@" "];
                            name = [name stringByAppendingString:lastname];
                            [duplicates addObject:name];
                        }
                    }
}
NSLog(@"Exited loop!");

NSLog的输出是：

Firstname = Bob
People count = 3
Exited loop!

我尝试用[people count]代替3，但输出相同。我没有休息;在我的代码中的任何地方声明，所以我不知道问题是什么......

Answer 1

您正在测试并递增内循环中的i变量;它应该是j：

for (int j = 0; j < [people count]; j++)

当像这样循环时，我通常使用另一种for-style循环：

for (Person *p in peopleArray) {
  // do something amazing here with p.
}

Answer 2

您正在测试并在内循环中递增i。

此外，内环应仅测试小于（或者大于）i的j的值。目前，您正在测试每对两次。

int n = [people count]; // minor perf improvement
for (int i = 1; i < n; ++i) {
    ...
    for (int j = 0; j < i; ++j) {
        ...

这也避免了j != i测试。

为了进一步改进，而不是内循环，将人名添加到NSSet并测试该集的成员身份以识别重复项。对于大量人群来说，这要快得多。

Answer 3

是的，for (int j = 0; i < [people count]; i++)

这是我们所犯的错误，比我们想承认的更多。

循环退出1个循环后

3 个答案: