c++ - gdb - 列出当前函数的源而不输入其名称

Question

在 GDB 中，命令:

list function

将列出该函数的所有源代码。

是否有一个命令可以列出您当前所在函数的所有源代码，而无需您手动输入函数名称？

Answer 1

(gdb) help list
List specified function or line.
With no argument, lists ten more lines after or around previous listing.
"list -" lists the ten lines before a previous ten-line listing.
One argument specifies a line, and ten lines are listed around that line.
Two arguments with comma between specify starting and ending lines to list.
Lines can be specified in these ways:
  LINENUM, to list around that line in current file,
  FILE:LINENUM, to list around that line in that file,
  FUNCTION, to list around beginning of that function,
  FILE:FUNCTION, to distinguish among like-named static functions.
  *ADDRESS, to list around the line containing that address.
With two args if one is empty it stands for ten lines away from the other arg.

*ADDRESS 很有趣。

在 x86/x64 上，当前指针位于 rip 寄存器中，因此:

(gdb) list *$pc
0x7ffff7b018a0 is at ../sysdeps/unix/syscall-template.S:82.
77  in ../sysdeps/unix/syscall-template.S

示例来自 cat 命令，因为我手头没有任何调试信息。