使用iOS 6 SDK和Xcode 4.5构建armv6架构（目标iOS 3.1.3）？ [英] Build for armv6 architecture (target iOS 3.1.3) with iOS 6 SDK and Xcode 4.5?

问题描述

是否有机会使用iOS 6 SDK开发针对iOS 3.1.3的应用程序？如果我使用旧的部署目标打开我的项目，编译器会警告我：

is there a chance to use the iOS 6 SDK to develop applications that target iOS 3.1.3? If I open my project with the old deployment target, the compiler warns me:

iOS deployment targets lower than 4.3 are not supported

这可能是因为显然只有 armv7 和 armv7s 可以构建二进制文件。如果我手动添加 armv6 （在iPhone 3G之前使用设备），也有几个警告：

This is probably due to the fact that apparently only armv7 and armv7s binaries can be built. If I add armv6 (working in devices until the iPhone 3G) manually, there are several warnings as well:

warning: no rule to process file '$(PROJECT_DIR)/main.m' of type sourcecode.c.objc for architecture armv6
...
...
warning: architecture armv6 is not supported (current ARCHS = "armv6 armv7 armv7s").

那我该如何处理呢？在使用iOS 6 SDK进行开发时，有没有办法支持iOS 3.1.3？如果没有，我最终将不得不转移到iOS 4.3的最低目标。甚至可以提交仅支持版本> = 4.3的新版本吗？我的用户在早期版本上会发生什么？他们是否根本没有被告知不兼容的更新，因此将继续使用我的应用程序的旧版本？如果没有，是否可以实现？

So how do I deal with this? Is there a way to support iOS 3.1.3 while developing with the iOS 6 SDK? If not, I will eventually have to move to the lowest target to be iOS 4.3. Is it even possible to submit a new version that supports only versions >= 4.3? What happens to my users on earlier versions? Are they simply not being informed about the incompatible update and hence going to continue to use the older version of my app? If not, can this be achieved?

推荐答案

使用Xcode 4.5（即iOS 6 SDK），您可以构建的最低版本是iOS 4.3。我想现在是3.1.3客户升级的时候了。

With Xcode 4.5 (ie iOS 6 SDK) the lowest version you can build for is iOS 4.3. I guess it is time for your 3.1.3 customers to upgrade.

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