For example in a videa about u-boot, https://www.youtube.com/watch?v=INWghYZH3hI, near time 43:01, I see the lecturer gives u-boot the kernel address and fdt address but not the initramfs address. (bootz 0x80000000 - 0x80800000) but linux boots to the login prompt and he can log in.
How come this is possible? I understand after the kernel boots it starts init process in the initramfs.(I forgot there were a precedence). Without initramfs, how is it possible to run login process or shell?
(it's related to programming so I ask it here. If requested I can move it to unix stackexchange. Is there a method of moving a question to somerewhere else automatically? guess not..)
this is what I learned from the comments with combination with my previous knowledge.
you can embed the initramfs.tar.gz file in the kernel binary image using CONFIG_INITRAMFS_SOURCE=nitramfs.cpio.gz in the configuration. (menuconfig). The initramfs image is placed at the end of the kernel image maybe (I remember).
But this was not the case in the youtube video I mentioned in my question. Near 40:29 in the video, the kernel boot command is shown to have "root=/dev/mmcblk0p1 rootfstype=ext4 rootwait console=tty0e,115200". So it's telling the kernel to use SD card 0's partition 1 as the root system after boot, instead of using initramfs.(when you want to use initramfs, you specify root=/dev/ram and pass the initramfs location. in qemu you use -initrd initramfs.cpio.gz option, or in real machine this information is passed through the device tree to the kernel, in the chosen node's initrd-start and initrd-end address.).