Shell Programming and Scripting

Hi,
I have a file a.lst which lists all files.
as
a.dat
b.dat
c.dat

I want to process these files mentioned in the list file in a loop.
Say I want to display only the first line of all the files a.dat , b.dat, c.dat.
How can I go about it?

Please help.

head -1 $(<a.lst)

hi ygor,

according to ur solution
the " head -1 a.lst " command willl give me the first
line of a.lst which will be a.dat
hwere as my problem is that i want to display the
first line of all the files (i.e a.dat, b.dat,c.dat ) listed in the a.lst file .
Can u get my problem ???????

ok, try it:

for i in `cat a.lst`
do
head -1 $i
done

By the way, you should be politely to who answered your question.

The standard output from a command enclosed in parenthesis preceded by a dollar sign or a pair of back single quotes can be used as part or all of a word; trailing new-lines are removed. In the second (archaic) form, the string between the quotes is processed for special quoting characters before the command is executed. The command substitution $(cat file) can be replaced by the equivalent but faster $(<file).

thanks tikual,
it is fine !!!
i just wanted to tell what the real problem was ????
sorry if framed my sentences in the earlier reply incorrectly ,
i am not a native english speaker !!
No intentions to cross fire nebody
Ne wayzzz thank u so much ........