UNIX for Dummies Questions & Answers

Hi!
I was trying to grep all the words in a wordlist, (twl), with no vowels. I had a hard time figuring out how to do it, but I finally lit on the -v flag. Here's my question:
Why does this work:

Code:

grep -v '[aeiou]' twl

But this doesn't:

Code:

grep '[^aeiou]' twl

In the second example, we're asking for lines which do not contain that class of letters, in this case the vowels, right? To my mind, it should come out the same. Clearly I'm missing something fundamental.

Read them like this:

grep lines without a 'vowel'

grep lines with a 'non-vowel'

Ok, I see. So in the second case, any line which contains a non-vowel returns 'true'. I guess we could use the -v flag there, and get vowels only.
Thanks!

Without -v:

Code:

grep '^[^aeiou]*$' twl

Regards,
Alister

It works, of course, but I would've thought the asterisk modifier, like some other things I tried, would return almost every line. Because if it goes through a line, and comes across a vowel, isn't that an instance of zero times, returning a "true"? Obviously, that's not what happens. I'm just not sure why. One of the ways I tried it, was with a "+" to say "at least one instance."
Also, I didn't think to "cap-off" the expression with a caret and dollar, but now I see why that's necessary.

Note that my pattern is anchored at both ends by ^ and $. That means that the entire line must consist exclusively of 0 or more non-vowels.

Regards,
Alister

grep uses REGEX.
In REGEX the * means the previous character can be repeated zero...many times.
A previous [^aeiou] is a 'non-vowel' character.
[^aeiou]* is any repetition of a 'non-vowel'.
The ^ at the beginning is 'beginning of line', and $ at the end means 'end of line'.
So you read the entire thing as

grep lines with any repetition of a 'non-vowel' between the beginning and the end of the line.

In other words: no other characters are allowed.