如何在 iOS 4 上禁用快速应用程序切换(多任务)?

问题描述

在 iOS 4 中，当用户按下 Home 按钮时，正在运行的应用程序将保存当前状态并进入后台.但是我正在开发一个应用程序，我不希望它进入后台.我希望它自杀，就像在 iOS 3 及之前的版本中一样.我该怎么做?谢谢.

In iOS 4, when a user presses the Home button, the running app will save the current state and go to background. However I'm developing an app and I don't want it to go to the background. I want it to kill itself, just like in iOS 3 and before. How can I do it? Thanks.

推荐答案

参见退出后台执行"library/ios/#documentation/iPhone/Conceptual/iPhoneOSProgrammingGuide/Introduction/Introduction.html%23//apple_ref/doc/uid/TP40007072-CH1-SW1" rel="noreferrer">iOS 应用程序编程指南:

See "Opting Out of Background Execution" in the iOS Application Programming Guide:

"...您可以明确选择退出通过添加后台执行模型UIApplicationExitsOnSuspend 键您的应用程序的 Info.plist 文件和将其值设置为 YES."

"...you can explicitly opt out of the background execution model by adding the UIApplicationExitsOnSuspend key to your application’s Info.plist file and setting its value to YES."

这篇关于如何在 iOS 4 上禁用快速应用程序切换(多任务)?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！