macos 如何使用 xcodebuild 构建特定架构？
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How can I build a specific architecture using xcodebuild?
提问by pqnet
I have legacy code that relies on pointers being 32-bitand want to use xCodeBuildto build that code from command line. This doesn't work for some reason. Here's the command I use:
我有依赖于指针的遗留代码，32-bit并希望用于xCodeBuild从command line. 由于某种原因，这不起作用。这是我使用的命令：
xcodebuild -configuration Debug -arch i386
-workspace MyProject.xcworkspace -scheme MyLib
here's the output I get
这是我得到的输出
[BEROR]No architectures to compile for
(ONLY_ACTIVE_ARCH=YES, active arch=x86_64, VALID_ARCHS=i386).
Clearly it's trying to build x86_64code and failing miserably since I only enabled i386from VALID_ARCHSin xCode project settings.
显然，它试图建立x86_64的代码和失败草草收场，因为我只启用i386从VALID_ARCHSXcode中的项目设置。
Is there a way to make it understand I don't want a 64-bitlibrary?
有没有办法让它明白我不想要一个64-bit图书馆？
回答by Laurent Etiemble
You have to set the ONLY_ACTIVE_ARCHto NOif you want xcodebuildto use the ARCHSparameters. By passing these parameters, you can force the proper architecture.
如果要使用参数ONLY_ACTIVE_ARCH，NO则必须设置为。通过传递这些参数，您可以强制使用正确的架构。xcodebuildARCHS
xcodebuild ARCHS=i386 ONLY_ACTIVE_ARCH=NO -configuration Debug -workspace MyProject.xcworkspace -scheme MyLib
See this referencefor details.
有关详细信息，请参阅此参考资料。
回答by yoAlex5
xcodebuild ONLY_ACTIVE_ARCH...
//or
Build Settings -> Build Active Architecture Only -> ONLY_ACTIVE_ARCH
YES- build binary with a single architecture for a connected device
YES- 为连接的设备构建具有单一架构的二进制文件
NO- build binary for a specific -arch(valid architecturesaka VALID_ARCHS) if it was specified or for all the architectures in other cases
NO- 为特定-arch（valid architectures又名VALID_ARCHS）构建二进制文件（如果已指定）或其他情况下的所有体系结构
The recommendation is to use Yesfor Debug(it save a build time) and No for Release build.
建议Yes用于 Debug（它节省构建时间）和 No 用于发布构建。
Note: it is safety to run on simulator
注意：在模拟器上运行是安全的
To check the version use lipo -info[About]
要检查版本使用lipo -info[关于]
