I have this regular expression \\..\\{3\\}$
I want to understand how this expression works to match a string. My thought is that it matches any 8 characters at the end of the line. Is that how this expression works?
If so, I think something like this would match the string:
rs.efg$tu
Assuming "typical" regexp's (sadly, different tools handle slightly different rexexps, and the GNU and POSIX versions are also different, and then there has been some version drift...), this parses as [Need Unicode-Art in markup...]
\. . \{3\} $
▲ ▲ ▲ ▲
│ │ │ │
│ │ │ └─ End of line
│ │ └─ Preceding exactly 3 times (the '\' makes '{' special...)
│ └─ Any character (except '\n')
└─ A literal '.' ('.' is special, '\' makes it un-special)
So this means a dot and 3 random characters before the end of the line.
Constructions like * or \{3\} (if the last one is even supported) apply to the last character, or the last parentesis (probably \( ... \), but that is again regexp-dialect-dependent). Check the manual for the exact tool you are using.
. is escaped, so it is a literal .. Good ASCII breakdown though.