If there is a better or more correct way to do this, anyone feel free to correct me, as my grep-fu is muy pobre. $ cat * | grep -c ' */' 6386 Doesn't that

If there is a better or more correct way to do this, anyone feel free to correct me, as my grep-fu is muy pobre. $ cat * | grep -c ' */' 6386 Doesn't that * in your regular expression mean "greedily match as many of the preceding SPACE character as possible including zero, followed by the /" ? So, what you've done is counted all the slashes in the code. Let's imagine that there are not many divide /, and there are not that many pathname /, so you're double counting the comments and should at least divide your 6386 in half you want $ cat * | grep -c ' \*/' the backslash is protected by the single quotes, so it will get passed to grep where it will mean "literally *" rather than the regular expression operator * (now I'm thinking my \ is going to get swallowed up by HN so I'd better double them? ed: nope, didn't have to double, but I did have to backslash escape the asterisk after literally)