从GDB文档：
你可以这样得到mystruct的类型：

tp = mystruct.type

并通过tp.fields()在字段上迭代
不需要邪恶的变通办法;- ）

**更新：**GDB 7.4刚刚发布。从announcement：

现在，结构和联合类型的类型对象允许使用标准Python字典（Map）方法访问字段。

邪恶的解决方法：

python print eval("dict(" + str(mystruct)[1:-2] + ")")

我不知道这是否是通用的。作为一个演示，我写了一个最小的例子test.cpp

# include <iostream>

struct mystruct
{
  int i;
  double x;
} mystruct_1;

int main ()
{
  mystruct_1.i = 2;
  mystruct_1.x = 1.242;
  std::cout << "Blarz";
  std::cout << std::endl;
}

现在，我像往常一样运行g++ -g test.cpp -o test，并启动gdb test。

(gdb) break main
Breakpoint 1 at 0x400898: file test.cpp, line 11.
(gdb) run
Starting program: ...

Breakpoint 1, main () at test.cpp:11
11        mystruct_1.i = 2;
(gdb) step
12        mystruct_1.x = 1.242;
(gdb) step
13        std::cout << "Blarz";
(gdb) python mystruct = gdb.parse_and_eval("mystruct_1")
(gdb) python print mystruct
{i = 2, x = 1.242}
(gdb) python print eval("dict(" + str(mystruct)[1:-2] + ")")
{'i': 2, 'x': 1.24}
(gdb) python print eval("dict(" + str(mystruct)[1:-2] + ")").keys()
['i', 'x']

这些天来：

(gdb) python import sys; print(sys.version)
3.10.8 (main, Oct 15 2022, 19:00:40)  [GCC 12.2.0 64 bit (AMD64)]

......它变得简单多了，属性是dict中的键-下面是一个示例：
test_struct.c：

# include <stdint.h>

# include <stdio.h>

struct mystruct_s {
  uint8_t member;
  uint8_t list[5];
};
typedef struct mystruct_s mystruct_t;

mystruct_t mystruct = {
  .member = 0,
  .list = { 10, 20, 30, 40, 50 },
};

int main(void) {
  printf("mystruct.member %d\n", mystruct.member);
  for(uint8_t ix=0; ix<sizeof(mystruct.list); ix++) {
    printf("mystruct.list[%d]: %d\n", ix, mystruct.list[ix]);
  }
}

然后编译并输入gdb：

gcc -g -o test_struct.exe test_struct.c
gdb --args ./test_struct.exe

...然后在gdb中：

(gdb) b main
Breakpoint 1 at 0x140001591: file test_struct.c, line 16.

(gdb) python ms = gdb.parse_and_eval("mystruct")
(gdb) python print(ms)
{member = 0 '\000', list = "\n\024\036(2"}
(gdb) python print(ms['member'])
0 '\000'
(gdb) python print(ms['list'])
"\n\024\036(2"

(gdb) python for ix in range(0,5): print("mystruct.list[{}]: {}".format(ix, ms['list'][ix]))
mystruct.list[0]: 10 '\n'
mystruct.list[1]: 20 '\024'
mystruct.list[2]: 30 '\036'
mystruct.list[3]: 40 '('
mystruct.list[4]: 50 '2'