如何创建 makefile 而不必包含项目中的每个源文件？
我正在上一些 C++ 课程，并向老师发送了 9 个练习，每个练习都是一个名为“ex$”的简单目录，其中 $ 是数字。每个目录都有一个名为“ex$.cpp”的源文件。我想创建一个 makefile，它允许我输入：
make ex$
它将构建一个与“ex$”目录中编译的源文件相对应的可执行文件。问题是我想做到这一点而不为每个练习创建目标（某种“通用目标”）。我还需要有一个“all”目标，该目标将进入以“ex”开头的每个目录并在那里构建可执行文件。我怎样才能做到这一点？
I'm taking some C++ classes and have send the teacher 9 exercises, each exercise is a simple directory with the name 'ex$' where $ is the number. Each directory has a single source file named 'ex$.cpp. I want to create a makefile that will allow me to type:
make ex$
And it will build a executable that corresponds to the compiled source file inside 'ex$' directory. The catch is that I want to do that without creating a target for each exercise(Some kind of 'generic target'). I also need to have an 'all' target that will go into each directory starting with 'ex' and build the executable there. How can I do that?
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如果您的所有 C++ 目标都可以使用基本相同的命令构建，那么您可以相当轻松地完成此操作。阅读本文。尤其要查找
$@。由于这是教育的一部分，所以我将把其余的内容模糊化。If all your C++ targets can be built with essentially the same command, you can do this fairly easily. Read this. Look for
$@, in particular. Since this is part of an education, I'll leave the rest vague.我还可以建议您查看 CMake，它会为您使用 IMO 生成更好的 makefile。初始的高学习曲线可实现主要的长期收益。 :)
Can I also suggest looking at CMake which will make better makefiles for you to use IMO. Initial high learning curve for major long term gain. :)