Question

给定一个定义了大量结构的C头文件和一个原始十六进制字符串，我想将字符串解析为相应的C结构。

我已经研究过struct.unpack（）来完成这个，但是我无法想出一种自动派生格式字符串的方法（这是可取的，因为这个头文件经常被更新）。

是否找到了为struct.unpack（）生成格式字符串的方法？或者是否有更简单的方法将十六进制字符串解析为C结构？

import struct

'''
Example structure:

typedef struct {
  struct {
    uint8_t   a_flag:1;
    uint8_t   b_flag:1;
    uint8_t   c_flag:1; 
    uint8_t   d_flag:1;
    uint8_t   unused:3;  
    uint8_t   e_flag:1; 
  } PACKED flag_byte_0;

  struct {
    uint8_t   unused:6;
    uint8_t   f_flag:1; 
    uint8_t   g_flag:1; 
  } PACKED flag_byte_1;

 struct {
    uint8_t   unused:5;
    uint8_t   h_flag:1; 
    uint8_t   i_flag:1;
    uint8_t   j_flag:1; 
 } PACKED flag_byte_2;

 struct {
    uint8_t   unused:2;
    uint8_t   k_flag;
    uint8_t   l_flag:1;
    uint8_t   unused_2:4;
 } PACKED flag_byte_3;

  uint16_t    field_a;
  uint16_t    field_b; 
  uint32_t    field_c:24;
  uint32_t    field_d:8; 
} PACKED struct_example; 
'''

if __name__ == '__main__':
    hex_string = '\x10\x08\x00\x00\x3d\x00\x08\xd7\x90\x00\x00\x0a'
    format_string = 'BBBBHHI'
    struct.unpack(format_string, hex_string)
    # returns (16, 8, 0, 0, 61, 55048, 167772304)
    # really want:
    #   a_flag:1
    #   g_flag:1
    #   ...
    #   field_a: 0x3d00
    #   etc...

Answer 1

使用cffi：

from cffi import FFI

ffi = FFI()
ffi.cdef('''
typedef struct {
  struct {
    uint8_t   a_flag:1;
    uint8_t   b_flag:1;
    uint8_t   c_flag:1;
    uint8_t   d_flag:1;
    uint8_t   unused:3;
    uint8_t   e_flag:1;
  } flag_byte_0;

  struct {
    uint8_t   unused:6;
    uint8_t   f_flag:1;
    uint8_t   g_flag:1;
  } flag_byte_1;

  struct {
    uint8_t   unused:5;
    uint8_t   h_flag:1;
    uint8_t   i_flag:1;
    uint8_t   j_flag:1;
  } flag_byte_2;

  struct {
    uint8_t   unused:2;
    uint8_t   k_flag:1;
    uint8_t   l_flag:1;
    uint8_t   unused_2:4;
  } flag_byte_3;

  uint16_t    field_a;
  uint16_t    field_b;
  uint32_t    field_c:24;
  uint32_t    field_d:8;
} struct_example;
''')


data = '\x10\x08\x00\x00\x3d\x00\x08\xd7\x90\x00\x00\x0a'
buf = ffi.new('char[]', data)
st = ffi.cast('struct_example*', buf)

print st.flag_byte_0.a_flag
print st.flag_byte_1.g_flag
print st.flag_byte_2.h_flag
print st.flag_byte_2.i_flag
print st.flag_byte_2.j_flag
print st.flag_byte_3.k_flag
print st.flag_byte_3.l_flag
print st.field_a
print st.field_b
print st.field_c
print st.field_d

将字节字符串解析为C结构

1 个答案: