I have been trying to debug my xv6 locally (It is a 32-bit code). But I am working on a 64 bit machine and my gdb is 64 bit. Whenever I type
make gdb
I encountered the following error
+ target remote localhost:26000
.gdbinit:23: Error in sourced command file:
localhost:26000: Connection timed out.
(gdb) target remote localhost:26000
localhost:26000: Connection timed out.
I have a .gdbinit file. I have the following line in it
target remote localhost:26000
It's returning an error at this line. I initially thought it is because of 32-bit architecture of my machine. But if I go to some other directory and type the same message in the gdb console I am receiving the same error. I can't actually understand why it should connect to localhost(i.e what is need of a server here? ) as GDB is a normal GNU debugger right? and what is the reason for its failure in the above case?
Thanks in advance
Your question doesn't make any sense (to me).
As far as I can tell, you are asking: "why does GDB try to connect to localhost:2600".
The answer is: because you asked GDB to do so with the target remote localhost:26000 entry in your ~/.gdbinit.
Update:
I am asking what does that command do?
It instructs GDB to perform remote debugging, and to connect to gdbserver listening on the given port.
In general, you should never put anything into your ~/.gdbinit (or any other initialization file) that you don't understand. Doing so is akin to to taking an unknown object (a gun), putting it to your head, and squeezing a little lever on the side (a trigger).
and executing it giving me the above error so can you please say what is the error? and how to resolve it
Just delete the command (and any other commands that you don't know the meaning of) from ~/.gdbinit.