Difference between echo `ls -l` and echo "`ls -l`" ?
Hi guys,
Been messing around with shell programming for a couple of days and I found something that was pretty odd in the behavior of the echo command. Below is an example-:
When I type the following in my /home directory from my lxterminal in Debian-:
Now this is what I expected. I know whatever is put in ` ` is suppose to interpreted as a command in the BASH echo. So, therefore this is the expected output of ls -l on my /home directory.
But it gets weird when I remove the " " (double quotes) from the echo statement.
If you look closely you will see that it is exactly identical to the output that is given above except for the fact that it contains no line breaks. My question is why ?
Why are the line-breaks removed when I remove the double quotes from echo ? Whatever is there in ` `(tild signs) is supposed to be interpreted as a command right ?
If removal of double quotes " " means that there is a change in the meaning of the echo, then please tell me what is the special meaning that double quotes has in echo ?
Is there any difference between echo "$SHELL" and echo $SHELL ? I know the output is same but since one has double quotes and one doesn't is there any difference in the way that BASH treats them internally ?
I am using Debian Wheezy LXDE with Bash 4.2.
Thanks.
Well, for starters -- it's a pretty silly thing to do, either way. Why do echo "`command`" when you could just do command ?
But the difference is, quotes prevent a string from splitting on whitespace. Spaces, tabs and newlines all count as whitespace. If you do VAR="a b c", and use $VAR without putting it into quotes, the shell will split it in three.
So, quoted, it ends up being:
Code:
echo "one
incredibly
long
string
crammed
into
a
single
argument
containing
actual,
real
line
breaks
printed
literally"
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