如何检测包含链中的第一个 Makefile？

Question

假设我有一个目录~/dir1/dir2/dir3/，该目录中有一个文件Makefile，其中包含内容

include ../Makefile

~/dir1/dir2/Makefile与~/dir1/dir2/dir3/Makefile的内容相同

如果允许我控制~/dir1/Makefile 的内容，并且make 在~/dir1/dir2/dir3/ 中被调用，我如何获取字符串~/dir1/dir2/dir3/Makefile（第一个包含文件）用作@987654332 中的变量@?

可重复的示例（根据 MadScientist 的要求）：

1. 任何 unix 系统
2. mkdir -p ~/dir1/dir2/dir3
3. cd ~/dir1/dir2/dir3
4. echo 'include ../Makefile' > Makefile
5. cp ./Makefile ../Makefile

现在我们在 ~/dir1/dir2/dir3/ 和 ~/dir1/dir2/ 中有一个具有相同声明的 Makefile，即包含父目录中的 Makefile 的包含声明。

如果在~/dir1/dir2/dir3 中调用了make，那么包含链最终会停在~/dir1/Makefile

如何找到此包含链中第一个文件的路径，即~/dir1/dir2/dir3/Makefile 以用作~/dir1/Makefile 中的变量？

Answer 1

要获取从 CLI 调用的第一个 make 文件的完整路径：

make_path_called_from_cli:=$(firstword $(foreach mk,${MAKEFILE_LIST},$(abspath ${mk})))

获取包含当前make文件的make文件的完整路径：

$(info Using GNUmake Special variable to obtain Every make file used...)

$(info ${MAKEFILE_LIST})

$(info Compute the full path for all of those make files...)
all_make_file_full_path:=$(foreach mk,${MAKEFILE_LIST},$(abspath ${mk}))

$(info ${all_make_file_full_path})

$(info The last one from the list is the current make file...)

current_make_file_path:=$(lastword ${all_make_file_full_path})
$(info ${current_make_file_path})

$(info Now we remove the current make file from the list of make...)
all_make_file_path_except_current:=$(filter-out ${current_make_file_path}, ${all_make_file_full_path})

$(info ${all_make_file_path_except_current})

# Now the last one from the list is the make file which included the current one
parent_make_file_path:=$(lastword ${all_make_file_path_except_current})

$(info We finally have the answer...)
$(info ${parent_make_file_path})