Typically, Objective-C class are abstract by convention only—if the author documents a class as abstract, just don't use it without subclassing it. There is no compile-time enforcement that prevents instantiation of an abstract class, however. In fact, there is nothing to stop a user from providing implementations of abstract methods via a category (i.e. at runtime). You can force a user to at least override certain methods by raising an exception in those methods implementation in your abstract class:

[NSException raise:NSInternalInconsistencyException 
            format:@"You must override %@ in a subclass", NSStringFromSelector(_cmd)];

如果您的方法返回一个值，那么使用起来会更容易一些

@throw [NSException exceptionWithName:NSInternalInconsistencyException
                               reason:[NSString stringWithFormat:@"You must override %@ in a subclass", NSStringFromSelector(_cmd)]
                             userInfo:nil];

这样就不需要从方法中添加return语句了。

如果抽象类实际上是一个接口(即没有具体的方法实现)，使用Objective-C协议是更合适的选择。

也许这种情况只会发生在开发阶段，所以这是可行的:

- (id)myMethodWithVar:(id)var {
   NSAssert(NO, @"You most override myMethodWithVar:");
   return nil;
}

你可以使用@Yar提出的方法(做了一些修改):

#define mustOverride() @throw [NSException exceptionWithName:NSInvalidArgumentException reason:[NSString stringWithFormat:@"%s must be overridden in a subclass/category", __PRETTY_FUNCTION__] userInfo:nil]
#define setMustOverride() NSLog(@"%@ - method not implemented", NSStringFromClass([self class])); mustOverride()

在这里你会得到这样的消息:

<Date> ProjectName[7921:1967092] <Class where method not implemented> - method not implemented
<Date> ProjectName[7921:1967092] *** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[<Base class (if inherited or same if not> <Method name>] must be overridden in a subclass/category'

或断言:

NSAssert(![self respondsToSelector:@selector(<MethodName>)], @"Not implemented");

在这种情况下，你会得到:

<Date> ProjectName[7926:1967491] *** Assertion failure in -[<Class Name> <Method name>], /Users/kirill/Documents/Projects/root/<ProjectName> Services/Classes/ViewControllers/YourClass:53

您也可以使用协议和其他解决方案-但这是最简单的解决方案之一。

事实上，Objective-C没有抽象类，但是您可以使用协议来达到同样的效果。下面是例子:

CustomProtocol.h

#import <Foundation/Foundation.h>

@protocol CustomProtocol <NSObject>
@required
- (void)methodA;
@optional
- (void)methodB;
@end

TestProtocol.h

#import <Foundation/Foundation.h>
#import "CustomProtocol.h"

@interface TestProtocol : NSObject <CustomProtocol>

@end

TestProtocol.m

#import "TestProtocol.h"

@implementation TestProtocol

- (void)methodA
{
  NSLog(@"methodA...");
}

- (void)methodB
{
  NSLog(@"methodB...");
}
@end

创建抽象类的简单示例

// Declare a protocol
@protocol AbcProtocol <NSObject>

-(void)fnOne;
-(void)fnTwo;

@optional

-(void)fnThree;

@end

// Abstract class
@interface AbstractAbc : NSObject<AbcProtocol>

@end

@implementation AbstractAbc

-(id)init{
    self = [super init];
    if (self) {
    }
    return self;
}

-(void)fnOne{
// Code
}

-(void)fnTwo{
// Code
}

@end

// Implementation class
@interface ImpAbc : AbstractAbc

@end

@implementation ImpAbc

-(id)init{
    self = [super init];
    if (self) {
    }
    return self;
}

// You may override it    
-(void)fnOne{
// Code
}
// You may override it
-(void)fnTwo{
// Code
}

-(void)fnThree{
// Code
}

@end

与其尝试创建抽象基类，不如考虑使用协议(类似于Java接口)。这允许您定义一组方法，然后接受符合协议的所有对象并实现这些方法。例如，我可以定义一个操作协议，然后有一个这样的函数:

- (void)performOperation:(id<Operation>)op
{
   // do something with operation
}

其中op可以是任何实现Operation协议的对象。

如果您需要抽象基类做的不仅仅是定义方法，那么您可以创建一个常规的Objective-C类并防止它被实例化。只需重写- (id)init函数，并使其返回nil或assert(false)。这不是一个非常干净的解决方案，但由于Objective-C是完全动态的，所以实际上没有与抽象基类直接等价的东西。