Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
342 views
in Technique[技术] by (71.8m points)

c - Difference in position-independent code: x86 vs x86-64

I was recently building a certain shared library (ELF) targeting x86-64 architecture, like this:

g++ -o binary.so -shared --no-undefined ... -lfoo -lbar

This failed with the following error:

relocation R_X86_64_32 against `a local symbol' can not be used when making a shared object; recompile with -fPIC

Of course, it means I need to rebuild it as position-independent code, so it's suitable for linking into a shared library.

But this works perfectly well on x86 with exactly the same build arguments. So the question is, how is relocation on x86 different from x86-64 and why don't I need to compile with -fPIC on the former?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

I have found a nice and detailed explanation, which boils down to:

  1. x86-64 uses IP-relative offset to load global data, x86-32 cannot, so it dereferences a global offset.
  2. IP-relative offset does not work for shared libraries, because global symbols can be overridden, so x86-64 breaks down when not built with PIC.
  3. If x86-64 built with PIC, the IP-relative offset dereference now yields a pointer to GOT entry, which is then dereferenced.
  4. x86-32, however, already uses a dereference of a global offset, so it is turned into GOT entry directly.

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...