问题描述
我在类中有一个NSInteger,如下
在@interface中:
code> NSInteger currentRow;
@property(assign)NSInteger currentRow;
在@implementation:
@synthesize currentRow;
做 [self setCurrentRow:0]
工作正常,但使用 [self currentRow]
只是由于某种原因返回null,不知道为什么。当使用断点时,我可以看到 currentRow
的值是 0
所以它已设置好,但我不能。
在Objective-C中,重要的是要区分对象和原始类型。
对象始终作为指针存储,它是对象在内存中的位置。指针只是一个数字。使用 NSLog
,您可以使用%p
查看此值。你可以在调试器中显示它,像这样: print myObject
。指针显示为十六进制数字,并带有 0x
前缀。 nil
实质上是位置零( 0x0000
)。当你分配任何类型的对象时,你会得到一个不为零的指针。当您将对象分配给变量时,您只需复制内存地址,而不是复制对象。使用 NSLog
,您可以使用%@
打印出对象的描述
。在调试器中,像这样: print-object myObject
。
原始类型 NSInteger
不是对象。而不是存储指针,通常你只是存储的值。当您分配一个 NSInteger
变量时,您将创建一个值的副本。您可以使用 print
在调试器中查看该值。或者像这样: NSLog(%ld,(long)currentRow)
。当您分配一个基元时,您复制其值。不要对原语使用%@
或 print-object
- 它们期望对象。
(我说通常你只是存储值,因为你也可以指向原始类型,在像你的情况下,这是没有必要的。)
[self currentRow]
返回0,就像你设置它。 (此外,因为Objective-C保证实例变量的初始化,即使你不设置它,它也会返回0)。
问题是,你期望一个指向一个对象的指针。如何修正程式码取决于您的使用方式:
- 如果您使用
print-object currentRow
,将其更改为打印currentRow
。 - 如果使用
NSLog(%@,currentRow)
,将其更改为NSLog(%ld,(long)currentRow)
。 - 如果你在其他地方使用
currentRow
,那么你的实例变量和属性类型改为NSNumber *
,一个对象类型。使用[self setCurrentRow:[NSNumber numberWithInt:0]]
设置。
I have an NSInteger in my class like so
In @interface:
NSInteger currentRow;
@property (assign) NSInteger currentRow;
In @implementation:
@synthesize currentRow;
Doing [self setCurrentRow:0]
seems to work fine, but using [self currentRow]
just returns null for some reason, not sure why. When using breakpoints I can see that the value for currentRow
is 0
so it has set fine, but I can't get the value back.
In Objective-C, it's important that you distinguish between objects and primitive types.
An object is always stored as a pointer, which is the object's location in memory. A pointer is just a number. With NSLog
, you can use %p
to see this value. You can display it in the debugger too, like this: print myObject
. A pointer is displayed as a hexadecimal number, with a 0x
prefix. nil
is essentially location zero (0x0000
). When you allocate any kind of object, you'll get a pointer which isn't zero. When you assign an object to a variable, you are simply copying the memory address, not duplicating the object. With NSLog
, you can use %@
to print out an object's description
. In the debugger, like this: print-object myObject
.
Primitive types like NSInteger
aren't objects. Instead of storing a pointer, usually you just store the value. When you assign an NSInteger
variable, you make a copy of the value. You can see the value in the debugger using print
. Or like this: NSLog("%ld", (long)currentRow)
. When you assign a primitive, you copy its value. Don't use %@
or print-object
with primitives — they expect objects.
(I say "usually you just store the value," because you can make pointers to primitive types, too. In situations like yours however it's not necessary.)
[self currentRow]
returns 0, just like you set it. (Furthermore, because Objective-C guarantees initialization of instance variables, it'll return 0 even if you don't set it.)
The problem is that you're expecting a pointer to an object. How you fix your code depends on how you're using it:
- If you're using
print-object currentRow
, change it toprint currentRow
. - If you're using
NSLog("%@", currentRow)
, change it toNSLog(%"ld", (long)currentRow)
. - If you're using
currentRow
somewhere else, where an object is required, change your instance variable and property types toNSNumber *
, an object type. Set it with[self setCurrentRow:[NSNumber numberWithInt:0]]
.
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