Hello, For anyone’s help.. I asked this question to unix stackexchage and received an answer. regards, Chan Kim From: Chan Kim <ckim@xxxxxxxxxx> Hello all, I'm learning u-boot these days and have a question. Using qemu model, I could (using qemu) load the “Image” (for arm64) on memory location I want and I could run the kernel booting using `booti ker_add initrd_addr dtb_addr` command. As you know in arm64, the uImage format is not supported. Then, when building linux kernel for arm (32 bit) using u-boot, what does “make LOADADDR=0x20008000 uImage” mean? Regardless of this LOADADDR, we can load (from SD card or network) the uImage anywhere we want using u-boot shell. Then what is this LOADADDR? Does it mean that when we run `bootm` command, the u-boot program relocates the kernel image inside the uImage to 0x20008000? (and the kernel built not using position independent, PC-relative addressing..using hard absolute addresses..) Some years ago, we used kernel start address during the build and I had to load the kernel image exactly on that address. But this time I could load ‘Image’ anywhere I want and could run it. Why? Is it because I (not knowingly) compiled the arm64 ‘Image’ using position independent address(-PIE option)? If someone with expertise could explain to me what’s happening here, I would really really appreciate it. Thank you. Best regards, Chan Kim |
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