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几天来我一直在努力解决这个问题，尝试了各种方法，但它们似乎都没有以我可以使用的方式工作......

问题。

我得到一个任意字节流。字节中隐藏着一些语义元素。有大括号、方括号和方括号。这些表示三个不同的东西 - {} 是一个否。字节范围，例如 {17} 是 17 个字节。[] 是字节值，例如 [90:95] 是字节 x90、x91、x92、x93、x94、x95。() 是字节值“或”选项，例如 (46|47) 表示 x46 或 x47。

我必须检测其他语法结构，“！”，“*”，“？” 和 ”：”。

示例字节流：524946(46|58){4}434452367672736E

我正在尝试过滤它，所以我得到了类似的东西：

1 string 524946
2 token (46|58)
3 token {4}
4 string 434452367672736E

一旦我把它分开，我就可以进一步处理它。

我最接近它的工作（它丑陋丑陋的代码......）： http: //pastebin.com/XLg2H0PW

我尝试了一些正则表达式，但我可以让它不将语法单元内的字符串字节计为普通字符串元素：

range_masks_list =  [(m_mask1.span()) for m_mask1 in re.finditer("\{([0-9]+|[0-9]+-[0-9]+|[0-9]+-\*)\}",sequence)] ## looks for {int}, {int-int} and {int-*}
byte_masks_list =  [(m_mask2.span()) for m_mask2 in re.finditer("\[[a-fA-F0-9]{2}:[a-fA-F0-9]{2}]",sequence)] ## looks for [a:b] where a and b are byte ranges
options_sets_list = [(m_mask3.span()) for m_mask3 in re.finditer("\(([a-fA-F0-9]{2})+\|([a-fA-F0-9]{2})+(\|([a-fA-F0-9]{2})+)*\)",sequence)] ## looks for regex or clauses e.g. (a|b)
string_chunk_list = [(m_mask4.span()) for m_mask4 in re.finditer("([a-fA-F0-9]{2})+",sequence)] ## looks for uninterrupted hex byte spans

这看起来像：

def do_fragmenter(self,sequence):
    """ converts the grep grammer normalised string into a set of fragments and offsets for sig population"""
    sequence = sequence.replace(" ","")
    range_masks_list =  [(m_mask1.span()) for m_mask1 in re.finditer("\{([0-9]+|[0-9]+-[0-9]+|[0-9]+-\*)\}",sequence)] ## looks for {int}, {int-int} and {int-*}
    byte_masks_list =  [(m_mask2.span()) for m_mask2 in re.finditer("\[[a-fA-F0-9]{2}:[a-fA-F0-9]{2}]",sequence)] ## looks for [a:b] where a and b are byte ranges
    options_sets_list = [(m_mask3.span()) for m_mask3 in re.finditer("\(([a-fA-F0-9]{2})+\|([a-fA-F0-9]{2})+(\|([a-fA-F0-9]{2})+)*\)",sequence)] ## looks for regex or clauses e.g. (a|b)
    string_chunk_list = [(m_mask4.span()) for m_mask4 in re.finditer("([a-fA-F0-9]{2})+",sequence)] ## looks for uninterupted hex byte spans 
    string_chunks = []
    string_chunks_len = []
    for pair in string_chunk_list:
        string_chunks.append(sequence[pair[0]:pair[1]])
        string_chunks_len.append(len(sequence[pair[0]:pair[1]]))
    print zip(string_chunks,string_chunks_len)
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1 回答 1

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只要考虑到您定义的语法元素，您可能会使用这样的东西（用您需要的处理替换打印）：

#! /usr/bin/python3.2

import re

a = '524946(46|58){4}434452[22:33]367672736E'
patterns = [ ('([0-9a-fA-F]+)', 'Sequence '),
    ('(\\([0-9a-fA-F]+\\|[0-9a-fA-F]+\\))', 'Option '),
    ('({[0-9a-fA-F]+})', 'Curly '),
    ('(\\[[0-9a-fA-F]+:[0-9a-fA-F]+\\])', 'Slice ') ]

while a:
    found = False
    for pattern, name in patterns:
        m = re.match (pattern, a)
        if m:
            m = m.groups () [0]
            print (name + m)
            a = a [len (m):]
            found = True
            break
    if not found: raise Exception ('Unrecognized sequence')

产量：

Sequence 524946
Option (46|58)
Curly {4}
Sequence 434452
Slice [22:33]
Sequence 367672736E
于 2013-02-18T23:30:47.093 回答