通过取消引用来更改 NSArray 中的值？

发布于 2024-08-12 08:14:41 字数 769 浏览 6 评论 0 原文

我遇到了与 Objective-C 中数组中的指针相关的问题。

我想做的是获取 NSArray 中的指针，将它们传递给一个方法，然后将返回的值分配回原始指针（属于数组的指针）。

根据我从 C 和 C++ 中了解到的信息，通过取消引用数组中的指针，我应该能够更改它们指向的值...这是我正在使用的代码，但它不起作用（值电话根据 NSLog 输出，指向永远不会改变）。

NSArray *phoneNumbers = [phoneEmailDict objectForKey:@"phone"];
    for (NSString* phone in phoneNumbers) {
        (*phone) = (*[self removeNonNumbers:phone]);
        NSLog(@"phone:%@", phone);
    }

下面是我将 NSString* 传递给的方法签名：

- (NSString*) removeNonNumbers: (NSString*) string;

如您所见，我使用变量phone 迭代phoneNumbers 中的每个NSString*。我将电话传递给removeNonNumbers:，它返回修改后的NSString*。 I 然后取消引用从removeNonNumber 返回的指针并将值分配给phone。

正如您所知，我可能不太了解 Objective-C 对象。我很确定这在 C++ 或 C 中可以工作，但我不明白为什么它在这里不起作用！预先感谢您的帮助！

原文

I've come across a problem related to pointers within arrays in objective-c.

What I'm trying to do is take the pointers within an NSArray, pass them to a method, and then assign the returned value back to the original pointer(the pointer which belongs to the array).

Based on what I know from C and C++, by dereferencing the pointers within the array, I should be able to change the values they point to... Here is the code I'm using, but it is not working (the value phone points to never changes based on the NSLog output).

NSArray *phoneNumbers = [phoneEmailDict objectForKey:@"phone"];
    for (NSString* phone in phoneNumbers) {
        (*phone) = (*[self removeNonNumbers:phone]);
        NSLog(@"phone:%@", phone);
    }

And here is the method signature I am passing the NSString* to:

- (NSString*) removeNonNumbers: (NSString*) string;

As you can see, I am iterating through each NSString* within phoneNumbers with the variable phone. I pass the phone to removeNonNumbers:, which returns the modified NSString*. I Then dereference the pointer returned from removeNonNumber and assign the value to phone.

As you can tell, I probably do not understand Objective-C objects that well. I'm pretty sure this would work in C++ or C, but I can't see why it doesn't work here! Thanks in advance for your help!

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

不回头走下去 2024-08-19 08:14:41

是的，那是行不通的。您需要一个 NSMutableArray：

NSMutableArray * phoneNumbers = [[phoneEmailDict objectForKey:@"phone"] mutableCopy];
for (NSUInteger i = 0; i < [phoneNumber count]; ++i) {
  NSString * phone = [phoneNumbers objectAtIndex:i];
  phone = [self removeNonNumbers:phone];
  [phoneNumbers replaceObjectAtIndex:i withObject:phone];
}
[phoneEmailDict setObject:phoneNumbers forKey:@"phone"];
[phoneNumbers release];

Yeah, that's not going to work. You'll need an NSMutableArray:

NSMutableArray * phoneNumbers = [[phoneEmailDict objectForKey:@"phone"] mutableCopy];
for (NSUInteger i = 0; i < [phoneNumber count]; ++i) {
  NSString * phone = [phoneNumbers objectAtIndex:i];
  phone = [self removeNonNumbers:phone];
  [phoneNumbers replaceObjectAtIndex:i withObject:phone];
}
[phoneEmailDict setObject:phoneNumbers forKey:@"phone"];
[phoneNumbers release];

回复收藏 0 原文

呆萌少年 2024-08-19 08:14:41

您无法取消引用 Objective-C 对象变量。它们始终是指针，但您应该将它们视为原子值。您需要改变数组本身以包含您正在生成的新对象。

回复收藏 0 原文

撩人痒 2024-08-19 08:14:41

NSArray 不是 C/C++ 风格的数组。它是一个 Objective-C 对象。您需要使用 NSArray 类的实例方法，用于对其执行操作。

在 Objective-C 中，您永远不会“取消引用”对象指针来设置其值。

另外，您正在使用所谓的快速枚举，不允许突变。

回复收藏 0 原文

疑心病 2024-08-19 08:14:41

您还可以使用 enumerateObjectsUsingBlock：。

NSArray *array = [NSArray array];
__block NSMutableArray *mutableCopyArray = [array mutableCopy];
[mutableCopyArray enumerateObjectsUsingBlock:^(id object, NSUInteger idx, BOOL *stop) {
    [mutableCopyArray replaceObjectAtIndex:idx withObject:[object modifiedObject]];
}];

查看如何迭代 NSArray？

You can also use enumerateObjectsUsingBlock:.

NSArray *array = [NSArray array];
__block NSMutableArray *mutableCopyArray = [array mutableCopy];
[mutableCopyArray enumerateObjectsUsingBlock:^(id object, NSUInteger idx, BOOL *stop) {
    [mutableCopyArray replaceObjectAtIndex:idx withObject:[object modifiedObject]];
}];

Checkout How do I iterate over an NSArray?

回复收藏 0 原文