Пример #1
0
 def __new__(cls, pattern):
     """
     If the pattern is completely static (no wildcards are present) a
     L{BytePattern} is created instead. That's because searching for a
     fixed byte pattern is faster than searching for a regular expression.
     """
     if '?' not in pattern:
         return BytePattern(HexInput.hexadecimal(pattern))
     return object.__new__(cls, pattern)
Пример #2
0
 def __new__(cls, pattern):
     """
     If the pattern is completely static (no wildcards are present) a
     L{BytePattern} is created instead. That's because searching for a
     fixed byte pattern is faster than searching for a regular expression.
     """
     if '?' not in pattern:
         return BytePattern( HexInput.hexadecimal(pattern) )
     return object.__new__(cls, pattern)
Пример #3
0
    def __init__(self, hexa):
        """
        Hex patterns must be in this form::
            "68 65 6c 6c 6f 20 77 6f 72 6c 64"  # "hello world"

        Spaces are optional. Capitalization of hex digits doesn't matter.
        This is exactly equivalent to the previous example::
            "68656C6C6F20776F726C64"            # "hello world"

        Wildcards are allowed, in the form of a C{?} sign in any hex digit::
            "5? 5? c3"          # pop register / pop register / ret
            "b8 ?? ?? ?? ??"    # mov eax, immediate value

        @type  hexa: str
        @param hexa: Pattern to search for.
        """
        maxLength = len([x
                         for x in hexa if x in "?0123456789ABCDEFabcdef"]) / 2
        super(HexPattern, self).__init__(HexInput.pattern(hexa),
                                         maxLength=maxLength)
Пример #4
0
    def __init__(self, hexa):
        """
        Hex patterns must be in this form::
            "68 65 6c 6c 6f 20 77 6f 72 6c 64"  # "hello world"

        Spaces are optional. Capitalization of hex digits doesn't matter.
        This is exactly equivalent to the previous example::
            "68656C6C6F20776F726C64"            # "hello world"

        Wildcards are allowed, in the form of a C{?} sign in any hex digit::
            "5? 5? c3"          # pop register / pop register / ret
            "b8 ?? ?? ?? ??"    # mov eax, immediate value

        @type  hexa: str
        @param hexa: Pattern to search for.
        """
        maxLength = len([x for x in hexa
                            if x in "?0123456789ABCDEFabcdef"]) / 2
        super(HexPattern, self).__init__(HexInput.pattern(hexa),
                                         maxLength = maxLength)