Why is ld.so a shared object?

Question

I'm currently writing a dynamic linker and I there is something bugging me. Why is ld.so (my system's dynamic linker) a shared object ? Why can't it just be a static executable (ET_EXEC) ?

I tried looking for answer in the linux kernel's binfmt_elf.c but to my understanding it clearly shows that your ELF interpreter could be a static executable.

EDIT:I think my thinking sums up to: Can a dynamic linker be a simple ELF executable (ET_EXEC) ?

binfmt_elf.c.559:

/* First of all, some simple consistency checks */
    if (interp_elf_ex->e_type != ET_EXEC &&
        interp_elf_ex->e_type != ET_DYN)
        goto out;

PS: I hope this is the right place, I don't know if I'm supposed to put it here or on Unix stack exchange. I'm also sorry if my question is stupid but not having an answer to this is driving me crazy.

Answer 1

Can a dynamic linker be a simple ELF executable (ET_EXEC) ?

Yes, it can.

However, an ET_EXEC must be loaded at the address it was linked at, and that address may conflict with the address at which a.out itself is linked. If such conflict happens, the kernel will either kill the process before it even starts, or it will mmap a.out "on top of" ld.so, and the resulting binary will crash.

You can move ld.so out of the way of usual a.out link address, but someone can always link a.out at un-usual address.

If instead you link ld.so as ET_DYN, with load address of zero, then none of the above problems could happen.