objdump, displaying without offsets

Question

When dumping an executable file, I only want the code segment to be printed on the standard output, not offsets and binary form of the code. I cannot achieve it from

man objdump

Is there a way?

Answer 1

You can suppress the object code hex dump with

--no-show-raw-insn

If you have jumps in the code then you need the offsets to make sense of them, but if you really want to strip them, filter the code with something like:

objdump -d --no-show-raw-insn myfile.o | perl -p -e 's/^\s+(\S+):\t//;'

Example output:

0000000000000000 <foo>:
retq
lea    0x0(%rsi),%rsi
lea    0x0(%rdi),%rdi
Disassembly of section .gnu.linkonce.t.goo:

0000000000000000 <goo>:
retq
lea    0x0(%rsi),%rsi
lea    0x0(%rdi),%rdi

Answer 2

If you want to achieve same output as Peeter Joot showed but without using Perl command line. Then you can use Grep and Cut tool instead, like shown below.

objdump -d --no-show-raw-insn my_binary_file.o | grep "^ " | cut -f2,3