为什么gcc用'self = [super initDesignatedInit]'来警告不兼容的结构赋值;''调用派生类？ [英] Why does gcc warn about incompatible struct assignment with a `self = [super initDesignatedInit];' call in derived class?

问题描述

  @interface ASCIICodeBase：NSObject {
 @protected 
 char code_ [4]; 
} 
  - （Base *）initWithASCIICode：（const char *）code; 
 @end 
 
 @implementation ASCIICodeBase 
  - （ASCIICodeBase *）initWithCode：（const char *）代码len：（size_t）len {
 if（len == 0 || len> 3）{
 return nil; 
} 
 if（self = [super init]）{
 memset（code_，0，4）; 
 strncpy（code_，code，3）; 
} 
返回自我; 
 
 @end 
 
 @interface CountryCode：ASCIICodeBase 
  - （CountryCode *）initWithCode：（const char *）code; 
 @end 
 @implementation CountryCode 
  - （CountryCode *）initWithCode：（const char *）code {
 size_t len = strlen（code）; 
 if（len！= 2）{
 return nil; 
} 
 self = [super initWithCode：code len：len]; //这里
返回自我; 
} 
 @end 
  

在标有here的行上，我得到以下gcc警告：

警告：不兼容Objective-C类型指定'struct ASCIICodeBase *'，预期'struct CurrencyCode *'


这段代码有什么问题，或者我应该有 ASCIICodeBase return ID ？或者在这里行上使用强制转换？

使用（id） c $ c>作为返回值类型。

Objective-C不支持协变声明。考虑：

  @interface NSArray：NSObject 
 +（id）array; 
 @end 
  

现在，您可以调用 + array NSArray 和 NSMutableArray 。前者返回一个不可变数组，后者返回一个可变数组。由于Objective-C缺乏协变声明支持，如果上面声明为返回（NSArray *），则子类方法的客户端必须转换为`（NSMutableArray * ）。丑陋，脆弱，容易出错。因此，使用泛型类型通常是最直接的解决方案。

因此...如果您声明一个返回特定类的实例的方法，则类型转换明确。如果你声明了一个将被覆盖的方法，并且覆盖可能返回一个子类和，它返回一个子类的事实将暴露给客户端，然后使用（id）。

指定的初始值设定项的工作方式相同。根据实现的上下文（当然）， -init * 方法可以返回几乎任何类型的实例。因此，返回类型的初始化方法是协变的，因此，您需要使用（id）作为返回类型。

无需提交错误 - 已有几个错误。

---

请注意，LLVM现在有一个 instancetype 关键字，可用于替换声明中的 id 像上面那样。这意味着这个方法有效地返回一个实例传递一个 isKindOfClass：测试类的实际名称。

I have the following base/derived class setup in Objective-C:

@interface ASCIICodeBase : NSObject {
 @protected
  char code_[4];
}
- (Base *)initWithASCIICode:(const char *)code;
@end

@implementation ASCIICodeBase
- (ASCIICodeBase *)initWithCode:(const char *)code len:(size_t)len {
  if (len == 0 || len > 3) {
    return nil;
  }
  if (self = [super init]) {
    memset(code_, 0, 4);
    strncpy(code_, code, 3);
  }
  return self;
}
@end

@interface CountryCode : ASCIICodeBase
- (CountryCode *)initWithCode:(const char *)code;
@end
@implementation CountryCode
- (CountryCode *)initWithCode:(const char *)code {
  size_t len = strlen(code);
  if (len != 2) {
    return nil;
  }
  self = [super initWithCode:code len:len]; // here
  return self;
}
@end

On the line marked "here", I get the following gcc warning:

warning: incompatible Objective-C types assigning 'struct ASCIICodeBase *', expected 'struct CurrencyCode *'

Is there something wrong with this code or should I have the ASCIICodeBase return id? Or maybe use a cast on the "here" line?

解决方案

Use (id) as the return value type.

Objective-C doesn't support covariant declarations. Consider:

@interface NSArray:NSObject
+ (id) array;
@end

Now, you can call +array on both NSArray and NSMutableArray. The former returns an immutable array and the latter a mutable array. Because of Objective-C's lack of covariant declaration support, if the above were declared as returning (NSArray*), clients of the subclasses method would have to cast to `(NSMutableArray*). Ugly, fragile, and error prone. Thus, using the generic type is, generally, the most straightforward solution.

So... if you are declaring a method that returns an instance of a specific class, typecast explicitly. If you are declaring a method that will be overridden and that override may return a subclass and the fact that it returns a subclass will be exposed to clients, then use (id).

Designated initializers work the same way. An -init* method may return an instance of just about any type, depending on context of implementation (of course). Thus, the return type of initialization methods is covariant and, as a result, you need to use (id) as the return type.

No need to file a bug -- there are several already.

---

Note that LLVM now has an instancetype keyword that can be used in place of id in a declaration like the above. It means "this method returns an instance that passes an isKindOfClass: test of the class upon which it was called", effectively.

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