grep doubt

[jordilin]

I have the following files:
hla.txt
hhla.txt
hhhhla.txt
Why when I type:

ls | grep h*
I get nothing. I thought that the regular expression * means zero or more matches of the preceding character, so I should get the three files. If I do
ls | grep h*la then, obviously I get the three files listed. Any good explanation??

[meng]

Why use grep at all? Just
ls h*
Here, * substitutes for any number of characters, and not just 'h' characters.

[jordilin]

Why use grep at all? Just
ls h*
Here, * substitutes for any number of characters, and not just 'h' characters.

Yes, you are right, but I'm learning regular expressions, and I would like to know why grep h* doesn't come up with any result.

[hod139]

With grep you have to escape the metacharacters (? not 100% sure what they are called) with a "\". So your command would become


ls | grep h\*la

[lloyd_b]

First off - "*" is a *shell* wildcard, NOT a regex wildcard. In regular expressions, the "*" operator is used to specify "zero or more repetitions of the preceding character".

If you want all lines that start with the letter "h", then the correct syntax is "grep ^h". The "^" specifies "start of line", so the regex translates to "all lines the consist of the sequence <start of line><letter h>"

May I suggest reading the man page for grep? It contains tons of useful information on regex's that can be used with grep.

Lloyd B.

[duff]

like lloyd said, the * character is being expanded by the shell. you can prevent this by using single tick marks:

# echo h*
hhhhla.txt hhla.txt hla.txt
# echo 'h*'
h*
so if you type the command
# ls | grep 'h*' you should get what you're looking for.

BTW, if you're going to try a compicated regex, you may need to pass the -e parameter. check the man page.

[amo-ej1]

When you use strace you can see how the program gets executed. e.g.



e@lap:~$ ls | strace grep e* 2>&1 | grep execve
execve("/bin/grep", ["grep", "ea", "eagle", "echo2.pdf", "echo3.pdf", "echo.pdf"], [/* 30 vars */]) = 0


'execve' is the systemcall to launch a new binary, the second argument contains the parameters passed. And in this example you clearly see how the e* got expanded to an array.