Я играю с lldb
и пытаюсь вызвать мою быструю статическую функцию.Мне удалось найти его детали в изображении, но я не знаю, как его вызвать и передать аргументы.
Мой вывод:
(lldb) image lookup -vs $S5project19ViewControllerUtilsC07setRootbC010storyboard13withAnimationySo12UIStoryboardC_SbtFZ
1 symbols match '$S5project19ViewControllerUtilsC07setRootbC010storyboard13withAnimationySo12UIStoryboardC_SbtFZ' in /Users/user/Library/Developer/CoreSimulator/Devices/DC7A5199-424E-4E38-A8A8-CB99C7D8CF82/data/Containers/Bundle/Application/BE7BB249-DB64-4228-B64E-EB430BCCE29E/project.app/project:
Address: project[0x00000002000fc970] (project.__TEXT.__text + 1027840)
Summary: project`static project.ViewControllerUtils.setRootViewController(storyboard: __C.UIStoryboard, withAnimation: Swift.Bool) -> () at ViewControllerUtils.swift:7
Module: file = "/Users/user/Library/Developer/CoreSimulator/Devices/DC7A5199-424E-4E38-A8A8-CB99C7D8CF82/data/Containers/Bundle/Application/BE7BB249-DB64-4228-B64E-EB430BCCE29E/project.app/project", arch = "x86_64"
CompileUnit: id = {0x00000000}, file = "/Users/user/mobile/company/iOS/project/project/Utils/ViewControllerUtils.swift", language = "swift"
Function: id = {0x7700000075}, name = "static project.ViewControllerUtils.setRootViewController(storyboard: __C.UIStoryboard, withAnimation: Swift.Bool) -> ()", mangled = "$S5project19ViewControllerUtilsC07setRootbC010storyboard13withAnimationySo12UIStoryboardC_SbtFZ", range = [0x000000010bd0c780-0x000000010bd0c82e)
FuncType: id = {0x7700000075}, byte-size = 8, decl = ViewControllerUtils.swift:7, compiler_type = "(UIKit.UIStoryboard, Swift.Bool) -> ()
"
Blocks: id = {0x7700000075}, range = [0x10bd0c780-0x10bd0c82e)
LineEntry: [0x000000010bd0c780-0x000000010bd0c79d): /Users/user/mobile/company/iOS/project/project/Utils/ViewControllerUtils.swift:7
Symbol: id = {0x00002a71}, range = [0x000000010ac0c780-0x000000010ac0c830), name="static project.ViewControllerUtils.setRootViewController(storyboard: __C.UIStoryboard, withAnimation: Swift.Bool) -> ()", mangled="$S5project19ViewControllerUtilsC07setRootbC010storyboard13withAnimationySo12UIStoryboardC_SbtFZ"
Variable: id = {0x6600000092}, name = "storyboard", type = "UIKit.UIStoryboard", location = DW_OP_fbreg(-16), decl = ViewControllerUtils.swift:7
Variable: id = {0x66000000a0}, name = "withAnimation", type = "Swift.Bool", location = DW_OP_fbreg(-24), decl = ViewControllerUtils.swift:7
Variable: id = {0x66000000ae}, name = "self", type = "@thick project.ViewControllerUtils.Type", location = DW_OP_fbreg(-32), decl = ViewControllerUtils.swift:7
Исходный код:
final class ViewControllerUtils {
static func setRootViewController(storyboard: UIStoryboard, withAnimation: Bool) {
setRootViewController(window: UIApplication.shared.keyWindow, storyboard: storyboard, withAnimation: withAnimation)
}
}
Я знаю, что мог бы переключиться на Swift
и вызвать что-то вроде этого:
expression -l swift -O -- ViewControllerUtils.setRootViewController(storyboard: UIKit.UIStoryboard(name: "Main", bundle: nil), withAnimation: true)
Но я хотел бы узнать, как вызвать функцию, используя ее адрес или символ безссылка на конкретный класс, как указано выше.