Makefile of xv6

Question

I am reading the code of xv6, and find it hard to read the Makefile. Could you tell me how the following statements work:

1. "CFLAGS += $(shell $(CC) -fno-stack-protector -E -x c /dev/null >/dev/null 2>&1 && echo -fno-stack-protector)"
2. "LDFLAGS += -m $(shell $(LD) -V | grep elf_i386 2>/dev/null)"
3. "xv6.img: 
bootblock kernel fs.img 

dd if=/dev/zero of=xv6.img count=10000

dd if=bootblock of=xv6.img conv=notrunc

dd if=kernel of=xv6.img seek=1 conv=notrunc"

And how to learn Makefile in details? Could you recommand some good books?

Thank you!

Answer 1

1. CFLAGS are the options ( like -fno-stack-protector -E etc ) you pass to your compiler CC. $(CC) will be replaced by the actual compiler. ie CC should be initialized before this with something like set CC=gcc.

2. LDFLAGS are the options ( to your linker LD. += is just like your += operator in C. It updates to the already existing value of CFLAGS and LDFLAGS

3. This line means that xv6.img is dependent on bootblock kernel fs.img. That is we are telling make that, in order to build xv6.img, you need to build bootblock kernel and fs.img first

You can learn about dd command here

Here is the complete encyclopedic guide to make and Makefiles

http://www.gnu.org/software/make/manual/make.html

Answer 2

To add to the answer, the dd command acts like a copy command essentially taking an input file[if] and copying it's contents to the output file [of]. Count is an indicator of the number of the blocks that are to be copied.

The code provided by you relates with the building of the xv6.img file which contains the bootloader, file system and the kernel. The code is essentially copying 10000 blocks of zeros from the /dev/zero file into xv6.img. This is followed by copying from the bootblock(created by bootasm.S and bootmain.c as seen by the Makefile target) into sector 0. This is followed by seeking one sector and then copying the kernel into the image file.