如何配置conanfile.py使用本地已克隆的源码文件

核心思路 #

默认情况下，Conan会将conanfile.py所在目录的本地文件作为项目源码，无需额外克隆操作。只需移除示例中用于远程拉取源码的配置，并确保本地源码目录结构正确即可。

具体步骤 #

1. 移除SCM配置（若存在）
   多数示例中的scm字段用于从远程仓库克隆源码，直接删除该部分即可避免不必要的克隆行为：

   # 删掉这类SCM配置
   # scm = {
   #     "type": "git",
   #     "url": "https://github.com/xxx/xxx.git",
   #     "revision": "master"
   # }
   

2. 确保源码目录结构正确
   你的源码文件（如CMakeLists.txt、src/、include/等）需与conanfile.py处于同一目录（或其子目录），Conan默认会扫描当前目录下的文件（排除Conan生成的build/、conan/等文件夹）。

3. 在build()方法中直接引用本地源码
   根据Conan版本不同，配置方式略有差异：

   Conan 1.x示例

   from conans import ConanFile, CMake
   
   class MyProjectConan(ConanFile):
       name = "my_project"
       version = "1.0.0"
       settings = "os", "compiler", "build_type", "arch"
       generators = "cmake"
   
       def build(self):
           cmake = CMake(self)
           # 直接指向本地的CMakeLists.txt所在路径
           cmake.configure(source_folder=".")
           cmake.build()
   

   Conan 2.x示例

   from conan import ConanFile
   from conan.tools.cmake import CMake, CMakeToolchain, CMakeDeps
   
   class MyProjectConan(ConanFile):
       name = "my_project"
       version = "1.0.0"
       settings = "os", "compiler", "build_type", "arch"
       generators = "CMakeDeps", "CMakeToolchain"
   
       def build(self):
           cmake = CMake(self)
           # Conan 2.x默认使用当前目录作为源码目录，无需额外指定
           # 若源码在子目录，可添加参数：cmake.configure(source_folder="./src")
           cmake.configure()
           cmake.build()
   

4. 可选：过滤需要包含的源码文件
   若不想将当前目录下所有文件都纳入构建，可通过配置指定要包含的文件/目录：

   Conan 1.x

   exports_sources = "CMakeLists.txt", "src/*", "include/*"
   

   Conan 2.x

   from conan.tools.files import copy
   
   def export_sources(self):
       copy(self, "CMakeLists.txt", self.recipe_folder, self.export_sources_folder)
       copy(self, "src/*", self.recipe_folder, self.export_sources_folder)
       copy(self, "include/*", self.recipe_folder, self.export_sources_folder)
   

内容的提问来源于stack exchange，提问作者montjet