Shell Programming and Scripting

This is extremely dangerous.

First, I assume you mean awk command in the generic sense, rather than command awk literally (which does something entirely different).

If the awk script fails for any reason, the contents of file will be destroyed. Depending on the operating system and shell you're using, echo "$var" may modify the contents of file if the expansion of $var contains any backslash (\) characters or starts with a minus-sign (-) character.

thanks for your explications, i'm not an expert as you, i use always the same OS (OSX) and the same shell (bash 3.2), in my case i think that $var don't modify the contents of file

message to MadeInGermany
I never work with bigger file text.

Thx

Hi protocomm,
Try the following script:

Code:

var=$(awk '1' < file);echo "$var" > file

on OS X using bash when file contains:

Code:

-n

I think you will find that it changes the contents from the three characters -, n, and <newline> to an empty file.

On some other operating systems, much more common file contents could be converted in ways (some obvious and some subtle) that would render working shell scripts or correct C source code contained in file into streams of syntax errors.

I try your line code and after the treatment and put

Code:

-n

in the file, the contents not change, the file is the same that before

Interesting...

Code:

bash-3.2$ bash --version
GNU bash, version 3.2.51(1)-release (x86_64-apple-darwin13)
Copyright (C) 2007 Free Software Foundation, Inc.
bash-3.2$ printf -- "-n\n" > file
bash-3.2$ ls -l file
-rw-r--r--  1 dwc  staff  3 May 13 09:45 file
bash-3.2$ var=$(awk '1' < file);echo "$var" > file
bash-3.2$ ls -l file
-rw-r--r--  1 dwc  staff  0 May 13 09:46 file
bash-3.2$

i have the same result that you demonstrate…
But if i put something after the -n\n , the file is not empty…

In fact i don't understand that you want to explain for me, sorry…, i don't understand the interest…
In which case i will have a file just with -n\n as content ?

The point is that in message #13 in this thread, you implied that

Code:

cp filename file.backup && \
awk 'awk_commands' file.backup > file && \
rm filename.backup
        and
var=$(awk 'awk_commands' < file);echo "$var" > file

are equivalent.

I want the people with less experience reading this thread to understand that although they may be equivalent in some cases on some systems; in cases of resource shortages, I/O errors, or power failures and (depending on the size of file, the contents of the file, and the version of echo you're using), some or all of the contents of file may be destroyed by the code shown in red while the code shown in green would preserve the original contents of file or atomically replace the contents of file with the results of any changes made by the awk script

(And, yes, I know that undetected hardware errors could destroy the contents of file with either method. But, we all hope that we don't have to worry about those cases too often.)