How do I print only the value of a variable in GDB?

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I have this script of GDB commands:

$ cat gdb_commands.txt
set pagination off
set logging file output.txt
set logging on
file stuff
b *0x80000014
run
echo ***DIFF THIS***\n
echo eax:
print $eax
echo ebx:
print $ebx
echo ecx:
print $ecx
echo edx:
print $edx
echo ***DIFF THIS END***\n
quit

If I run it in GDB I get this:

$ gdb -q -x gdb_commands.txt
Breakpoint 1 at 0x80000014

Breakpoint 1, 0x80000014 in _start ()
***DIFF THIS***
eax:$1 = 1
ebx:$2 = 2
ecx:$3 = 3
edx:$4 = 4
***DIFF THIS END***
A debugging session is active.

    Inferior 1 [process 8947] will be killed.

Quit anyway? (y or n) [answered Y; input not from terminal]

So there is that ugly dollar sign thing. I can sed it out, but I would like to have GDB to do that. Is it possible?

(The reason I'm use GDB like this is because we are writing an emulator and want to test if it behaves correctly.)

gdb
asked on Stack Overflow Sep 4, 2013 by Calmarius • edited Aug 13, 2019 by Peter Mortensen

2 Answers

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ugly dollar sign thing ... I would like to have gdb to do that

You can control GDB's output precisely with printf command:

(gdb) print/x $rax
$1 = 0x7ffff7ffe2a0

(gdb) printf "0x%lx\n", $rax
0x7ffff7ffe2a0
answered on Stack Overflow Sep 5, 2013 by Employed Russian
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There is a command that does exactly that:

(gdb) help output
Like "print" but don't put in value history and don't print newline.
This is useful in user-defined commands.

output prints the variables without the $1 = and the newline.

answered on Stack Overflow Oct 30, 2020 by tartaruga_casco_mole

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