Omitting results containing '//' from grep searches

Question

I'm trying to search a substantial C++ codebase for a single line that is outputing text to the terminal that I'd like to suppress. I have grepped for std::cout and have had around 40 different files returned. The problems I face are:

• I did not add this myself so I have no idea where it is.
• It is a pointer only outputing memory location so I have no context in which to search for it.
• The codebase is enourmous and contains a great many other instances of sdt::cout that have once been used for debugging purposes and have since been commented out.

My question pertains to the last one. I am using

grep -rle 'std::cout' .

to search, which will return positive for instances of std::cout, //std::cout, // std::cout and any other occurence of std::cout sitting on a line that is actually commented out.

How can I modify my grep to omit any line containing // so I can eliminate the commented lines?

Answer 1

egrep -r '^([^/]/?)*std::cout' .

Answer 2

You could look for std::cout only when it isn't preceded by //. Regexp syntax does not include negation; every regexp can be negated by writing out its complement, but the complement can grow huge. Here, it's not too big:

grep -rl -E '^/?([^/]/?)*std::cout' .

You can also list all occurrences of std::cout and filter away the occurrences of //.*std::cout, but note that this will hide things like std::cout << foo; // std::cout << bar;.

grep -r 'std::cout' . | grep -vE '^[^:]*:.*//.*std::cout' | sed -e 's/:.*//'

Alternatively, you can run a tool that parses the C++ code, such as ctags.

Answer 3

grep -r 'std::cout' . | grep -v '^[ \t]*//'

will only omit the commented out lines, and not all the lines that have a comment on them.