无法访问返回对象的属性？

//SomeObject.h 
#import <Foundation/Foundation.h> 


@interface SomeObject : NSObject { 

} 
@property NSInteger aProperty; 
@end 

//main.m 

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

int main (int argc, const char * argv[]) 
{ 

    NSAutoreleasePool * pool = [[NSAutoreleasePool alloc] init]; 

    NSMutableDictionary *dictionary = [[NSMutableDictionary alloc] init]; 
    [dictionary setObject:[[[SomeObject alloc] init] autorelease] forKey:@"key1"]; 
    [dictionary setObject:[[[SomeObject alloc] init] autorelease] forKey:@"key2"]; 
    [dictionary objectForKey:@"key1"].aProperty = 5; //Error HERE 
    [dictionary release]; 

    [pool drain]; 
    return 0; 
}

但在该行的XCode给我这些错误：无法访问返回对象的属性？

error: Semantic Issue: Member reference type 'struct objc_object *' is a pointer; maybe you meant to use '->'? 
error: Semantic Issue: No member named 'aProperty' in 'struct objc_object'

我不能访问返回的对象的属性？（我的意思是，没有直接调用setter方法）

来源

2011-08-29 Michele De Pascalis

煤焦与指针“*”？我是noob，但我从来没有看到带有星号的字符，就像你做的那样“char * argv []” – doNotCheckMyBlog

这就像编写char argv [] []或char ** argv ...这是一个二元数组，每个标准C main接收命令行参数应该有这样的参数列表。 –

你需要将返回的对象：

((SomeObject*)[dictionary objectForKey:@"key1"]).aProperty = 5;

或：

[(SomeObject*)[dictionary objectForKey:@"key1"] setAProperty: 5];

或：

SomeObject* obj = [dictionary objectForKey:@"key1"]; 
obj.aProperty = 5;

来源

2011-08-29 10:34:43 Felix

仍然认为我不必在C++中做到这一点......谢谢！ –

稍等片刻，解决方案1应该是'（（SomeObject *）[dictionary objectForKey：@“key1”]）。aProperty = 5;' –

没关系，只是改正了它 – Felix

无法访问返回对象的属性？

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