为什么在指定NSStrings时快速枚举不跳过NSNumbers? [英] Why does fast enumeration not skip the NSNumbers when I specify NSStrings?

问题描述

我以为我知道如何使用快速枚举，但是对此我有些不了解.如果我创建三个NSString对象和三个NSNumber对象并将它们放在NSMutableArray中:

I thought that I knew how to use fast enumeration, but there is something I don't understand about it. If I create three NSString objects and three NSNumber objects and put them in an NSMutableArray:

NSString *str1 = @"str1";
NSString *str2 = @"str2";
NSString *str3 = @"str3";

NSNumber *nb1 = [NSNumber numberWithInt:1];
NSNumber *nb2 = [NSNumber numberWithInt:2];
NSNumber *nb3 = [NSNumber numberWithInt:3];

NSArray *array = [[NSArray alloc] initWithObjects:str1, str2, str3, nb1, nb2, nb3, nil];

然后我对所有NSString对象进行快速枚举，如下所示:

then I make do fast enumeration on all NSString objects, like this:

for (NSString *str in array) {
    NSLog(@"str : %@", str);
}

在控制台中，我得到以下结果:

In the console, I get this result :

2011-08-02 13:53:12.873 FastEnumeration[14172:b603] str : str1
2011-08-02 13:53:12.874 FastEnumeration[14172:b603] str : str2
2011-08-02 13:53:12.875 FastEnumeration[14172:b603] str : str3
2011-08-02 13:53:12.875 FastEnumeration[14172:b603] str : 1
2011-08-02 13:53:12.876 FastEnumeration[14172:b603] str : 2
2011-08-02 13:53:12.876 FastEnumeration[14172:b603] str : 3

我只记录了NSString，但是我得到了数组中每个对象的一行，甚至是NSNumber，我也不明白为什么.快速枚举是否总是使用数组中包含的每个对象?

I logged only the NSStrings, but I get a line for every object in the array, even the NSNumbers and I don't understand why. Does fast enumeration always use every object contained in an array?

推荐答案

当您编写这样的forin循环时，它会将数组中的每个对象强制转换为NSString，然后根据要求将其打印出来.

When you write a forin loop like that, it casts every object in the array as an NSString, then prints them out as requested.

如果只需要NSString，则需要编写如下内容:

If you want only the NSStrings, you would need to write something like this:

for (id obj in array) {
    if ([obj isKindOfClass:[NSString class]]) {
        NSLog(@"str: %@", obj);
    }
}

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