Question

我试图用swift调用一个objective-c方法，并且得到了这个奇怪的错误：

Cannot convert the expression's type 'Void' to type 'String!'

Swift代码：

XNGAPIClient.sharedClient().putUpdateGeoLocationForUserID("me",
        accuracy: 3000,
        latitude: location.coordinate.latitude as CGFloat,
        longitude: location.coordinate.longitude as CGFloat,
        ttl: 420, success: { (JSON: AnyObject!) in },
        failure: { (error: NSError!) in })

Objective-C方法：

- (void)putUpdateGeoLocationForUserID:(NSString*)userID
                             accuracy:(CGFloat)accuracy
                              latitude:(CGFloat)latitude
                             longitude:(CGFloat)longitude
                                   ttl:(NSUInteger)ttl
                               success:(void (^)(id JSON))success
                               failure:(void (^)(NSError *error))failure

如果我将所有内容转换为建议的类型：

XNGAPIClient.sharedClient().putUpdateGeoLocationForUserID("me" as String,
        accuracy: 3000 as CGFloat,
        latitude: location.coordinate.latitude as CGFloat,
        longitude: location.coordinate.longitude as CGFloat,
        ttl: 420 as Int,
        success: { (JSON: AnyObject!) in },
        failure: { (error: NSError!) in })

我收到以下错误：Cannot convert the expression's type 'Void' to type 'StringLiteralConvertible'

Answer 1

您的问题是location.coordinate.latitude和location.coordinate.longitude参数。例如，如果我将这些参数设置为Int，我可以重现您的问题。所以，试试：

XNGAPIClient.sharedClient().putUpdateGeoLocationForUserID("me" as String,
    accuracy: 3000 as CGFloat,
    latitude: CGFloat(location.coordinate.latitude),
    longitude: CGFloat(location.coordinate.longitude),
    ttl: 420 as Int,
    success: { (JSON: AnyObject!) in },
    failure: { (error: NSError!) in })

...也就是说，使用CGFloat构造函数而不是向下转换as来获取这些参数。（我猜想有一个聪明的东西在幕后为3000文字看起来应该是一个Int，否则那个可能不会工作，或者......）

我还提出了一个与Apple无关的错误消息的错误。我在使用错误的参数类型调用Objective C时看到了一些。

无法转换表达式＆＃39; Void＆＃39;输入＆＃39; String！＆＃39;

1 个答案: