如何通过构造解析一个bin文件?
how to parse a bin file by construct?
文件格式为:
from construct import *
file_format=Struct(
'n_links'/ Int32ul, # number of links
'links'/ Array(this.n_links, Int32ul), # links
'n_Items' / ??? # how to do here ?
)
02000000 0C000000 10000000 55000000 AA000000
n_links link_0 link_1 value_of_link_0 value_of_link_1
如何解析'n_Items'?
我解决了:
file_format=Struct(
'n_links'/ Int32ul, # number of links
'links'/ Array(this.n_links, Int32ul), # links
'n_Items' / Pointer(this.links[0], Int32ul[this.n_links])
)
t=file_format.parse('\x02\x00\x00\x00\x04\x00\x00\x00\x08\x00\x00\x00') #ok parse
文件格式为:
from construct import *
file_format=Struct(
'n_links'/ Int32ul, # number of links
'links'/ Array(this.n_links, Int32ul), # links
'n_Items' / ??? # how to do here ?
)
02000000 0C000000 10000000 55000000 AA000000
n_links link_0 link_1 value_of_link_0 value_of_link_1
如何解析'n_Items'?
我解决了:
file_format=Struct(
'n_links'/ Int32ul, # number of links
'links'/ Array(this.n_links, Int32ul), # links
'n_Items' / Pointer(this.links[0], Int32ul[this.n_links])
)
t=file_format.parse('\x02\x00\x00\x00\x04\x00\x00\x00\x08\x00\x00\x00') #ok parse