gdb doesnot display source code

[vijayanand_sodadasi]

I compiled a c program using


gcc -Wall -g example.c

I tried to debug the program using gdb.


gdb -tui a.out

But the source code of my program is not being displayed in the window.
How can I view the source code while I debug the program?

Thanks and Regards,
Vijayanand.

[oakz]

see the LIST command

also, read "info gdb" for all of the commands available to you from the (gdb) prompt

[jerome bettis]

apt-get DDD, it's a gui frontend to gdb and a lot easier to work with. gdb doesn't show enough info at once and is tedious to use.

[twopoint718]

It actually does this in two ways. Here's how I usually use GDB. After I compile my program with the '-g' flag (this inserts debugging information) I run this (if my program is the default 'a.out'):

$ gdb a.out

Then gdb prints a little about itself and dumps you into the command line. Here you can set a few break points, something like "b 19" for break on line 19 or "b foo" to break on function foo.

If you want to see a quick listing of the lines that are around the line you're on type:

(gdb) list

if you want to see a separate panel that always shows your code type:

C-x C-a (that is type and hold "Ctrl" while also pressing 'x', then hold "Ctrl" and press 'a').

This will show you a pane at the top that displays your code (GDB calls this "TUI" mode: GDB user manual (http://sourceware.org/gdb/current/onlinedocs/gdb_24.html#SEC253)). If you haven't already, type "run" and gdb will run your code. You'll see the top pane paused at your breakpoint (if you set one) or the last line that your program got to.

I hope this is more of what you were looking for.