Perl一线式操作include目录

Question

I have several hundred files with include directives that need to be changed.

The prototype is #include "MyFile.hpp" , but I need to append a directory in front of every include file like this: #include "project-head/MyFile.hpp" .

I've written a one-line perl script as follows:

perl -p -i -e 's|/(#include /")/(.*/.hpp)|/1project-head/2|g' test_file.hpp

But the test file doesn't seem to be changing. What is wrong with my with my perl expression?

Answer 1

You have multiple odd slashes in your Perl script. The backrefs should be \\1 , \\2 etc and many of the rest should be dropped entirely.

perl -p -i -e 's|(#include ")(.*\.hpp)|\1project-head/\2|g' test_file.hpp

In case it's not blindingly obvious, / and \\ are two different characters with radically different semantics.

The prototype for the substitution command is s/from/to/ but you can use, and have in your code used, different delimiters, so it's s|from|to| with no slashes at all (the only remaining forward slash is the directory separator between project-head and the original file name).

The backslash is commonly used as an escape character; for example, \\. matches a literal period (whereas an unescaped . is a regex metacharacter which matches any character except newline). It also has a special meaning in \\1 to refer back to the text matched by the first parenthesized subexpression (though in the replacement part $1 is preferred).