makefile 里的 sed 命令，里面的$*是 makefile 的一个自动变量吗

makefile 正在学习中，看到一个规则里面用了 sed 命令，sed 命令里用了$*，感觉只用正则表达式来解释$解释为行尾解释不通。

ldd@CD:~/hello$ gcc -M hello.c hello.o: hello.c /usr/include/stdc-predef.h /usr/include/stdio.h \ /usr/include/x86_64-linux-gnu/bits/libc-header-start.h \ /usr/include/features.h /usr/include/x86_64-linux-gnu/sys/cdefs.h \ /usr/include/x86_64-linux-gnu/bits/wordsize.h \ /usr/include/x86_64-linux-gnu/bits/long-double.h \ /usr/include/x86_64-linux-gnu/gnu/stubs.h \ /usr/include/x86_64-linux-gnu/gnu/stubs-64.h \ /usr/lib/gcc/x86_64-linux-gnu/9/include/stddef.h \ /usr/lib/gcc/x86_64-linux-gnu/9/include/stdarg.h \ /usr/include/x86_64-linux-gnu/bits/types.h \ /usr/include/x86_64-linux-gnu/bits/timesize.h \ /usr/include/x86_64-linux-gnu/bits/typesizes.h \ /usr/include/x86_64-linux-gnu/bits/time64.h \ /usr/include/x86_64-linux-gnu/bits/types/__fpos_t.h \ /usr/include/x86_64-linux-gnu/bits/types/__mbstate_t.h \ /usr/include/x86_64-linux-gnu/bits/types/__fpos64_t.h \ /usr/include/x86_64-linux-gnu/bits/types/__FILE.h \ /usr/include/x86_64-linux-gnu/bits/types/FILE.h \ /usr/include/x86_64-linux-gnu/bits/types/struct_FILE.h \ /usr/include/x86_64-linux-gnu/bits/stdio_lim.h \ /usr/include/x86_64-linux-gnu/bits/sys_errlist.h 

另外随便写了一个 hellowoald ，看了 gcc -M 的输出如上。

另外，为啥一定要用这个 d 文件，感觉有点没理解它的妙处？

PS：脚本方面比较弱，正在学习中，大佬们帮忙解答下。