如何只在gdb中打印变量的值？

Question

所以我有这个gdb命令的脚本：

$ 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

如果我运行它，我得到这个：

$ 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]

所以有一个丑陋的美元符号的东西。我可以sed它，但我想让gdb这样做。有可能吗？

（我之所以使用gdb是因为我们正在编写模拟器，并且想测试它是否正常运行。）

Answer 1

  丑陋的美元符号......我想让gdb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

Answer 2

有一个命令可以做到这一点：

(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显示不包含$1 = 和换行符的变量。