UNIX for Advanced & Expert Users

dear all,

can anyone clarify on this below regular expression output...because i feel this is not the output i should get according to the concepts...

NOTE :Executed on linux system

Code:

$ cat grepf
abbbbd
a.c
abc
ac
abcd
yyyy
$ grep ^\a.c  grepf
a.c
abc
abcd
$ grep a.c  grepf
a.c
abc
abcd
$ grep a\.c  grepf
a.c
abc
abcd

Sure...

Your use of \. didn't reach grep, because the shell saw the \ and escaped the . which means grep saw a.c and not a\.c

What you want to do is either

Code:

$ grep 'a.c' grepf

or

Code:

$ grep a\\.c grepf

It's pretty safe to say that since most shells do some kind of regular expression handling, when you use grep you should put the regexp in quotes.

Then wht is the meaning of "." and "\."

To the shell?

In a shell, . is used for globbing, which is similar but not the same as what grep does. Globbing is what lets you 'ls *' and get anything useful. The shell uses backslashes to mean to "take the next character literally", also much like in grep. But since the shell does command separation and grep does not it's useful in other places. For example if you have a filename with spaces, you can specify it like filename\ with\ spaces to avoid needing single or double quotes.

Thanks for ur timely reply...