如何使用 xcodebuild 构建特定架构? [英] How can I build a specific architecture using xcodebuild?

问题描述

我有依赖于 32 位 指针的遗留代码，并希望使用 xCodeBuild 从 命令行 构建该代码.由于某种原因，这不起作用.这是我使用的命令:

I have legacy code that relies on pointers being 32-bit and want to use xCodeBuild to build that code from command line. This doesn't work for some reason. Here's the command I use:

xcodebuild -configuration Debug -arch i386 
  -workspace MyProject.xcworkspace -scheme MyLib

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).

显然它正在尝试构建 x86_64 代码并且失败了，因为我只在 xCode 项目设置中从 VALID_ARCHS 启用了 i386.

Clearly it's trying to build x86_64 code and failing miserably since I only enabled i386 from VALID_ARCHS in xCode project settings.

有没有办法让它明白我不想要一个 64 位 库?

Is there a way to make it understand I don't want a 64-bit library?

推荐答案

如果你想要 xcodebuild，你必须将 ONLY_ACTIVE_ARCH 设置为 NO使用 ARCHS 参数.通过传递这些参数，您可以强制使用正确的架构.

You have to set the ONLY_ACTIVE_ARCH to NO if you want xcodebuild to use the ARCHS parameters. By passing these parameters, you can force the proper architecture.

xcodebuild ARCHS=i386 ONLY_ACTIVE_ARCH=NO -configuration Debug -workspace MyProject.xcworkspace -scheme MyLib

请参阅此参考资料 了解详情.

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