Assume the pattern length = 500 (you can change it to any value). By default this will create 20280 probabilities max.
pattern_create = ('Aa0'..'Zz9').to_a.join.each_char.first(500).joinIn case you need longer a pattern (ex. 30000) you can do the following
pattern_create = ('Aa0'..'Zz9').to_a.join
pattern_create = pattern_create * (30000 / 20280.to_f).ceilI'll assume the pattern was equal or less than “20280” and we are looking for “9Ak0” pattern characters. The pattern_create should be initialized from above
pattern_offset = pattern_create.enum_for(:scan , '9Ak0').map {Regexp.last_match.begin(0)}Note: This does not consider the Little-endian format, for that there is extra code that should be written. For more info, please take a look at the following [code][1].
puts (0..255).map {|b| ('\x%02X' % b)}{% hint style="warning" %}
- To change value presentation from (
\xeato0xea), change\x%xto0x%x - To make all letters capital (
\xeato\xEA) , change\x%xto\x%X{% endhint %}
(32..126).map {|c| c.chr}short and unclean
(32..126).map &:chrResources