Question

我的代码正在编译并运行良好，直到我升级到Xcode 6。

定义显示警告：自动属性合成不会合成属性＆＃39; hash＆＃39;因为它是'readwrite＆＃39;但它将被合成＆＃39; readonly＆＃39;通过另一个属性

@property (nonatomic, strong)               NSString       *hash;       // (get/compute) hash code of the place (master hash of images)

每当我访问_hash时，实现都会显示错误：使用未声明的标识符＆＃39; _hash＆＃39;

-(NSString *)hash {
    if (_hash) return _hash;

    // If place id, take it as the hash code
    NSString *poiID = self.info[@"id"];
    if (poiID) {
        _hash = [NSString stringWithFormat:@"id-%lu",(unsigned long)[self.address hash]];
    }
    else if (CLLocationCoordinate2DIsValid(self.location.coordinate)) {
        NSString *seed = [NSString stringWithFormat:@"%f,%f", self.location.coordinate.latitude, self.location.coordinate.longitude];
        _hash = [NSString stringWithFormat:@"location-%lu",(unsigned long)[seed hash]];
    }
    else if (self.address) {
        NSString *seed = self.address;
        _hash = [NSString stringWithFormat:@"address-%lu",(unsigned long)[seed hash]];
    }
    else {
        _hash = @"POI-unknownIDLocationOrAddress";
    }

    return _hash;
}

Answer 1

它没有编译，因为hash已经是NSObject的一部分了：

请参阅： https://developer.apple.com/library/ios/documentation/Cocoa/Reference/Foundation/Protocols/NSObject_Protocol/index.html#//apple_ref/occ/intfm/NSObject/hash

Answer 2

您需要添加以下行以自动生成setter方法：

@synthesize hash = _hash;

如果您不想要一个setter方法并且只想要一个只读属性：

@property (nonatomic, strong, readonly) NSString *hash;

Xcode 6：为什么这段代码现在不编译？

2 个答案: