如何仅打印GDB中变量的值？

我有以下GDB命令脚本：

$ 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

如果我在GDB中运行它，我会得到以下结果：

$ 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]

这就是丑陋的美元符号。我可以

sed

out，但我想让GDB来做这件事。可能吗

（我之所以这样使用GDB，是因为我们正在编写一个模拟器，并希望测试它的行为是否正确。）

丑陋的美元符号。。。我希望gdb能做到这一点

您可以使用

printf

命令精确控制GDB的输出：

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

(gdb) printf "0x%lx\n", $rax
0x7ffff7ffe2a0

---

有一个命令正好执行以下操作：

(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

打印不带

$1=

和换行符的变量