关于 -(NSDictionary)dictionaryWithObjectsAndKeys：和

Question

我有一个非常有趣的问题。

在我的一个类中，我声明了一个非常简单的实例方法 -(NSDictionary)dictionary; 这是以这种方式实现的：

- (NSDictionary *)dictionary {
    return [NSDictionary dictionaryWithObjectsAndKeys:
                              self.userID, @"id",
                              self.userName, @"name",
                              self.userSurname, @"sName",
                              self.userNickname, @"nName",
                              self.userPassword, @"pwd",
                              [NSNumber numberWithDouble:[self.userBirthday timeIntervalSince1970] * 1000], @"birthday",
                              self.userSex, @"sex",
                              self.userEmail, @"email",
                              self.userLanguage, @"locale",
                              [NSNumber numberWithLongLong:self.userCoin], @"coin",
                              self.userRate, @"rate",
                              [NSNumber numberWithDouble:self.userCoordinate.latitude], @"lat",
                              [NSNumber numberWithDouble:self.userCoordinate.longitude], @"lon",
                              self.userPlaces, @"userPlaces",
                              nil];
    }

声明此方法后，我的返回字典中没有 @"userPlaces" 键（self.userPlace 显然已被评估并且充满了对象）。

所以我改变了一点我的方法，如下所示：

- (NSDictionary *)dictionary {      
    NSMutableDictionary *toReturn = [NSMutableDictionary dictionary];
    [toReturn setValue:self.userID forKey:@"id"];
    [toReturn setValue:self.userName forKey:@"name"];
    [toReturn setValue:self.userSurname forKey:@"sName"];
    [toReturn setValue:self.userNickname forKey:@"nName"];
    [toReturn setValue:self.userPassword forKey:@"pwd"];
    [toReturn setValue:[NSNumber numberWithDouble:[self.userBirthday timeIntervalSince1970] * 1000] forKey:@"birthday"];
    [toReturn setValue:self.userSex forKey:@"sex"];
    [toReturn setValue:self.userEmail forKey:@"email"];
    [toReturn setValue:self.userLanguage forKey:@"locale"];
    [toReturn setValue:[NSNumber numberWithLongLong:self.userCoin] forKey:@"coin"];
    [toReturn setValue:self.userRate forKey:@"rate"];
    [toReturn setValue:[NSNumber numberWithDouble:self.userCoordinate.latitude] forKey:@"lat"];
    [toReturn setValue:[NSNumber numberWithDouble:self.userCoordinate.longitude] forKey:@"lon"];
    [toReturn setValue:self.userPlaces forKey:@"userPlaces"];

    return [NSDictionary dictionaryWithDictionary:toReturn];
}

所有密钥现在都存在于输出字典中！

这个问题让我疯狂地想要理解，但我的问题是...... 有什么原因导致第二种方法比第一种方法效果更好？

我没有找到原因。

Answer 1

[NSDictionary DictionaryWithObjectsAndKeys: 当发现 nil 值时停止添加对象。

因此，您尝试在第一个代码中添加的对象很可能是 nil。

Answer 2

类似于下面的代码的东西解决了我的问题，但有同样的例外：

(NSMutableDictionary *)dictionary {
return [NSMutableDictionary dictionaryWithObjectsAndKeys:
                          self.userID, @"id",
                          self.userName, @"name",
                          self.userSurname, @"sName",
                          self.userNickname, @"nName",
                          self.userPassword, @"pwd",
                          [NSNumber numberWithDouble:[self.userBirthday timeIntervalSince1970] * 1000], @"birthday",
                          self.userSex, @"sex",
                          self.userEmail, @"email",
                          self.userLanguage, @"locale",
                          [NSNumber numberWithLongLong:self.userCoin], @"coin",
                          self.userRate, @"rate",
                          [NSNumber numberWithDouble:self.userCoordinate.latitude], @"lat",
                          [NSNumber numberWithDouble:self.userCoordinate.longitude], @"lon",
                          self.userPlaces, @"userPlaces",
                          nil];
}