我正在尝试实现基于应答How to implement method swizzling swift 3.0?的swift 3中的方法
这是我的密码:
// MARK: - Swizzling
private let swizzling: (UIView.Type) -> () = { view in
let originalSelector = #selector(view.awakeFromNib)
let swizzledSelector = #selector(view.swizzled_localization_awakeFromNib)
let originalMethod = class_getInstanceMethod(view, originalSelector)
let swizzledMethod = class_getInstanceMethod(view, swizzledSelector)
method_exchangeImplementations(originalMethod, swizzledMethod)
}
extension UIView {
open override class func initialize() {
guard self === UIView.self else {
return
}
swizzling(self)
}
func swizzled_localization_awakeFromNib() {
swizzled_localization_awakeFromNib()
if let localizableView = self as? Localizable {
localizableView.localize()
}
}
}但在应用程序启动时,它就有理由崩溃了:
'-UINavigationController swizzled_localization_awakeFromNib:未识别的选择器发送到实例0x7fc7c8820400‘
我不明白为什么swizzled_localization_awakeFromNib打电话给UINavigationController。我在obj-c项目中使用这个,这是原因吗?它在斯威夫特2通过dispatch_once很好地工作。
我在swizzling(self),之前尝试了place断点,它在UIView上调用了一次,就像预期的那样。
发布于 2017-02-19 09:33:15
问题是awakeFromNib是NSObject的一种方法,而不是UIView的方法。您的代码使用一个UIView方法对UIView方法进行处理,当UINavigationController (或者不是UIView子类的NSObject的任何其他子类)上调用swizzled方法时,调用原始方法就会崩溃。
解决方案是尝试添加带有原始名称的swizzled方法(如http://nshipster.com/method-swizzling/中所述):
private let swizzling: (UIView.Type) -> () = { view in
let originalSelector = #selector(view.awakeFromNib)
let swizzledSelector = #selector(view.swizzled_localization_awakeFromNib)
let originalMethod = class_getInstanceMethod(view, originalSelector)
let swizzledMethod = class_getInstanceMethod(view, swizzledSelector)
let didAddMethod = class_addMethod(view, originalSelector, method_getImplementation(swizzledMethod), method_getTypeEncoding(swizzledMethod))
if didAddMethod {
class_replaceMethod(view, swizzledSelector, method_getImplementation(originalMethod), method_getTypeEncoding(originalMethod))
} else {
method_exchangeImplementations(originalMethod, swizzledMethod)
}
}https://stackoverflow.com/questions/42325078
复制相似问题