UNIX for Dummies Questions & Answers

because one backslah is not important in csh
needed X2

both of '\' and 'n' escape its special means
then
\\ --> \
\\ --> n

5 backslah also same state ( last of \ like newline char in shells and doesnt anything in command in below)

Code:

# echo RECORDDELIMITER \
? \\\\n
RECORDDELIMITER \n

Code:

# echo RECORDDELIMITER \\\\\n
RECORDDELIMITER \n

actually

Code:

\n

is special character it has a mean

Code:

# eval echo RECORDDELIMITER '\A'
RECORDDELIMITER A

eval(actually csh) trying process but \A has no special mean on csh
it will be process only echo and one of backslahs isnt here

Code:

# eval echo RECORDDELIMITER '\\A'
RECORDDELIMITER \A

And '\' and 'n' is a char in csh

Code:

# eval echo RECORDDELIMITER '\\ \n'
RECORDDELIMITER \ n

Regards
ygemici

I'm very sorry, I didn't mention the fact that I was using bash. So, the examples u gave don't give me the results that u mention. My question is :

Code:

eval echo \\h
eval echo \\\h

both give output as h, but

Code:

eval echo \\\\h

gives output as \h

Does eval here just remove backslashes (I know the use of eval in the case where substitutions occur)

* First of all let look the below

Code:

# echo \h\h\h\h
hhhh

\ --> does no effect to echo


* Anyway

Code:

 
# echo \++
++

first \ char is meaningless in here and echo after \ chars so \++

so echo already write on the newline

Code:

  \ is same --> echo \

Code:

 
# echo A
A
# echo \A
A

but the trick for use ""
use this to remove its special meanings

Code:

# echo "A"
A
# echo "\A"
\A

Anymore \ is calculating with "" too

Code:

 
# echo \\\\\++
\\++

\ and \\\\++
so echo \\\\++ --> is to be \\++
\\ \\ --> two backslah is mean (one backslah) + two backslah is mean (one backslah) --> one + one = two backslah ( \\ )

Code:

 
# echo "\\\\\++"
\\\++

"\" and \\\\++
so echo \\\\++ --> is to be \\\++
\\ \\ --> two backslah is mean (one backslah) + two backslah is mean (one backslah) --> one + one = two backslah ( \\ )
\\ plus "\" --> \\\ three backslah and ++ ---> \\\++





Now in examples

1-)

Code:

 
# echo \\++
\++

\ and \++
first \ char doesnt change in the output
so echo \++

2 -)

Code:

 
# echo \\\++
\++

\ and \\++
first \char .....
so echo \\++ --> is to be \++
because two \\ originally is one \
\\ --> \

3-)

Code:

 
[root@XXXx ~]# echo \\\\++ 
\\++

\ and \\\++
first \char .....
so echo \\\++ --> is to be \\++
\\ \ --> two backslah is mean (one backslah)+ one backslah --> one + one = two backslah ( \\ )

4-)

Code:

 
[root@XXXx ~]# echo \\\\\++
\\++

\ and \\\\++
so echo \\\\++ --> is to be \\++
\\ \\ --> two backslah is mean (one backslah) + two backslah is mean (one backslah) --> one + one = two backslah ( \\ )


* Eval ; so eval removes the first backslash..

Code:

 
[root@XXXx ~]# eval echo \\++ 
++

---> eval removes the backslash and echo processing..

Code:

 
[root@XXXx ~]# eval echo \\\++ 
++

---> eval removes the backslash and echo processing..

Code:

 
[root@XXXx ~]# eval echo \\\\++ 
\++

---> eval removes the backslash and echo processing..



Regards
ygemici

Thanks a gigaton ! It really was of great help. I will just mention what I have understood.

A backslash removes any special meaning of the symbol succeeding it. If there is no special symbol (like $,\,`,etc.) succeeding it, the backslash is removed.

Code:

 # echo \t
# t

In the following, the special meaning of the succeeding character is removed.

Code:

# set he is a boy
# echo $1
(Here special meaning of $ is retained)
# echo \$1
# $1 (here the special meaning is removed)

Similarly, in

Code:

echo \\\t

the first \ tells that the next slash is to be taken literally (and is itself not printed), so one slash (the second one) is printed. The third slash has no special character after it, so its not printed. The output is :

Code:

# \t

Similarly

Code:

# echo \\\\t
# \\t

. The first and 3rd \ tell that the next \ has to be taken literally, and hence the 2nd and 4th are printed.

As for a backslash within quotes, it has special meaning only when followed by \,$,',",newline. Otherwise, its a simple character.

Code:

# echo "\jao"
# \jao

Here, since \ is not followed by the above mentioned characters, it is treated as a simple character.

Code:

# echo "\\jao"
# \jao

Here, \ is followed by \ (a character in the above list) and so is treated specially (i.e. it makes the following character an ordinary one and is itself removed)

Now, for eval. Eval parses the argument once without executing the command, and then again, this time executing it too.

Code:

# eval echo \\g
# g

On the first parse, the first backslash makes the second \ be treated as an ordinary character, and is itself removed. So, after the first parse, the argument is \g. A similar thing happens in the second parse, and the output is therefore, g.

Code:

# eval echo \\\\g
# \g

Now, in the first parse, what happens is similar to the example I gave a few lines above (the same thing as u type it without eval), and so after the first parse u have : \\g as the argument. Now, in the second pass, the first / makes the next slash appear as an ordinary character, and is itself rmeoved, so we have a single slash followed by g as the answer.

I hope I have understood correctly.