Linux – How to single step ARM assembly in GDB on QEMU

armassemblygdblinuxqemu

I'm trying to learn about ARM assembler programming using the GNU assembler. I've setup my PC with QEmu and have a Debian ARM-HF chroot environment.

If I assemble and link my test program:

.text
.global _start
_start:
        mov     r0, #6
        bx      lr

with:

as test.s -o test.o
ld test.o -o test

Then load the file into gdb and set a breakpoint on _start:

root@Latitude-E6420:/root# gdb test
GNU gdb (GDB) 7.6.1 (Debian 7.6.1-1)
Copyright (C) 2013 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later 
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.  Type "show copying"
and "show warranty" for details.
This GDB was configured as "arm-linux-gnueabihf".
For bug reporting instructions, please see:
...
Reading symbols from /root/test...(no debugging symbols found)...done.
(gdb) break _start
Breakpoint 1 at 0x8054
(gdb)

How do I single step the code, display the assembler source code and monitor the registers?
I tried some basic commands and they did not work:

(gdb) break _start
Breakpoint 1 at 0x8054
(gdb) info regi
The program has no registers now.
(gdb) stepi
The program is not being run.
(gdb) disas
No frame selected.
(gdb) r
Starting program: /root/test 
qemu: Unsupported syscall: 26
qemu: uncaught target signal 11 (Segmentation fault) - core dumped
qemu: Unsupported syscall: 26
During startup program terminated with signal SIGSEGV, Segmentation fault.
(gdb)

Best Answer

Your problem here is that you're trying to run an ARM gdb under QEMU's user-mode emulation. QEMU doesn't support the ptrace syscall (that's what syscall number 26 is), so this is never going to work.

What you need to do is run your test binary under QEMU with the QEMU options to enable QEMU's own builtin gdb stub which will listen on a TCP port. Then you can run a gdb compiled to run on your host system but with support for ARM targets, and tell that to connect to the TCP port.

(Emulating ptrace within QEMU is technically very tricky, and it would not provide much extra functionality that you can't already achieve via the QEMU builtin gdbstub. It's very unlikely it'll ever be implemented.)

Related Solutions

Show current assembly instruction in GDB

You can switch to assembly layout in GDB:

(gdb) layout asm

See here for more information. The current assembly instruction will be shown in assembler window.

   ┌───────────────────────────────────────────────────────────────────────────┐
   │0x7ffff740d756 <__libc_start_main+214>  mov    0x39670b(%rip),%rax        #│
   │0x7ffff740d75d <__libc_start_main+221>  mov    0x8(%rsp),%rsi              │
   │0x7ffff740d762 <__libc_start_main+226>  mov    0x14(%rsp),%edi             │
   │0x7ffff740d766 <__libc_start_main+230>  mov    (%rax),%rdx                 │
   │0x7ffff740d769 <__libc_start_main+233>  callq  *0x18(%rsp)                 │
  >│0x7ffff740d76d <__libc_start_main+237>  mov    %eax,%edi                   │
   │0x7ffff740d76f <__libc_start_main+239>  callq  0x7ffff7427970 <exit>       │
   │0x7ffff740d774 <__libc_start_main+244>  xor    %edx,%edx                   │
   │0x7ffff740d776 <__libc_start_main+246>  jmpq   0x7ffff740d6b9 <__libc_start│
   │0x7ffff740d77b <__libc_start_main+251>  mov    0x39ca2e(%rip),%rax        #│
   │0x7ffff740d782 <__libc_start_main+258>  ror    $0x11,%rax                  │
   │0x7ffff740d786 <__libc_start_main+262>  xor    %fs:0x30,%rax               │
   │0x7ffff740d78f <__libc_start_main+271>  callq  *%rax                       │
   └───────────────────────────────────────────────────────────────────────────┘
multi-thre process 3718 In: __libc_start_main     Line: ??   PC: 0x7ffff740d76d
#3  0x00007ffff7466eb5 in _IO_do_write () from /lib/x86_64-linux-gnu/libc.so.6
#4  0x00007ffff74671ff in _IO_file_overflow ()
   from /lib/x86_64-linux-gnu/libc.so.6
#5  0x0000000000408756 in ?? ()
#6  0x0000000000403980 in ?? ()
#7  0x00007ffff740d76d in __libc_start_main ()
   from /lib/x86_64-linux-gnu/libc.so.6
(gdb)

Linux – How to change permissions for a folder and its subfolders/files in one step

The other answers are correct, in that chmod -R 755 will set these permissions to all files and subfolders in the tree. But why on earth would you want to? It might make sense for the directories, but why set the execute bit on all the files?

I suspect what you really want to do is set the directories to 755 and either leave the files alone or set them to 644. For this, you can use the find command. For example:

To change all the directories to 755 (drwxr-xr-x):

find /opt/lampp/htdocs -type d -exec chmod 755 {} \;

To change all the files to 644 (-rw-r--r--):

find /opt/lampp/htdocs -type f -exec chmod 644 {} \;

Some splainin': ^{(thanks @tobbez)}

chmod 755 {} specifies the command that will be executed by find for each directory
chmod 644 {} specifies the command that will be executed by find for each file
{} is replaced by the path
; the semicolon tells find that this is the end of the command it's supposed to execute
\; the semicolon is escaped, otherwise it would be interpreted by the shell instead of find

Best Answer

Related Solutions

Show current assembly instruction in GDB

Linux – How to change permissions for a folder and its subfolders/files in one step

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