这个答案对我有用
https://stackoverflow.com/a/35788694但它需要一些改编,比如创建一个桥接头来导入一些特定的IOKit部件.
首先,将IOKit.framework添加到项目中(单击“链接的框架和库”中的“”).
然后创建一个新的空“.m”文件,无论其名称如何.然后Xcode将询问它是否应该制作“桥接头”.说是的.
忽略“.m”文件.在Xcode刚刚创建的新“YOURAPPNAME-Bridging-Header.h”文件中,添加以下行:
#include <IOKit/IOKitLib.h> #include <IOKit/usb/IoUSBLib.h> #include <IOKit/hid/IOHIDKeys.h>
class USBDetector { class func monitorUSBEvent() { var portIterator: io_iterator_t = 0 let matchingDict = IOServiceMatching(kIoUSBDeviceClassName) let gNotifyPort: IONotificationPortRef = IONotificationPortCreate(kIOMasterPortDefault) let runLoopSource: Unmanaged<CFRunLoopSource>! = IONotificationPortGetRunLoopSource(gNotifyPort) let gRunLoop: CFRunLoop! = CFRunLoopGetCurrent() CFRunLoopAddSource(gRunLoop,runLoopSource.takeRetainedValue(),kcfRunLoopDefaultMode) let observer = UnsafeMutablePointer<Void>(unsafeAddressOf(self)) _ = IOServiceAddMatchingNotification(gNotifyPort,kIOMatchednotification,matchingDict,deviceAdded,observer,&portIterator) deviceAdded(nil,iterator: portIterator) _ = IOServiceAddMatchingNotification(gNotifyPort,kIOTerminatednotification,deviceRemoved,&portIterator) deviceRemoved(nil,iterator: portIterator) } } func deviceAdded(refCon: UnsafeMutablePointer<Void>,iterator: io_iterator_t) { var kr: kern_return_t = KERN_FAILURE while case let usbDevice = IOIteratorNext(iterator) where usbDevice != 0 { let deviceNameAsCFString = UnsafeMutablePointer<io_name_t>.alloc(1) defer {deviceNameAsCFString.dealloc(1)} kr = IORegistryEntryGetName(usbDevice,UnsafeMutablePointer(deviceNameAsCFString)) if kr != KERN_SUCCESS { deviceNameAsCFString.memory.0 = 0 } let deviceName = String.fromCString(UnsafePointer(deviceNameAsCFString)) print("Active device: \(deviceName!)") IOObjectRelease(usbDevice) } } func deviceRemoved(refCon: UnsafeMutablePointer<Void>,iterator: io_iterator_t) { // ... }
注意:deviceAdded和deviceRemoved需要是函数(而不是方法).
要使用此代码,只需启动观察者:
USBDetector.monitorUSBEvent()
这将列出当前插入的设备,并在每个新的USB设备插入/拔出事件上,它将打印设备名称.
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 [email protected] 举报,一经查实,本站将立刻删除。