Re: [Help-bash] Reading and handling "control" characters from a file

[Davide Brini]

On Sat, 21 Apr 2012 18:41:05 -0500, "Conrad J. Sabatier" <address@hidden>
wrote:

> On Sun, 22 Apr 2012 00:50:25 +0200
> Davide Brini <address@hidden> wrote:
> 
> > On Sat, 21 Apr 2012 14:13:52 -0400, Chet Ramey <address@hidden>
> > wrote:
> > 
> > > On 4/19/12 6:11 PM, Conrad J. Sabatier wrote:
> > > > I've just started doing a little prototyping in bash for a
> > > > program I'll eventually code most likely in C, and have hit a
> > > > serious stumbling block re: the handling of characters (bytes) in
> > > > the very low range of the ASCII table.
> > > 
> > > Bash variables are strings of characters.  There is a difference
> > > between a character with ASCII value 4 and a character with the
> > > ASCII value 52 ("4").  Shell arithmetic convers the latter into
> > > numbers using the equivalent of strtol() before use.
> > > 
> > > To make what you want work, you'll have to figure out some way to
> > > offset the value you read from the file (i.e., c+'0') before
> > > attempting to use it in an arithmetic context.  Maybe perl or
> > > something like that could help. It's quite difficult to do using
> > > just what the shell provides.
> > 
> > Shouldn't the single quote trick work here? eg
> > 
> > $ a=$(printf '\x4\n')
> > $ printf "%d\n" "'$a"
> > 4
> > 
> 
> Yes, I thought that was the solution for a brief time, but it still
> fails when the character is a newline, for some odd reason.  Bash seems
> to be rather fickle when it comes to certain things, allowing this,
> objecting to that.

You must be doing something odd, because newline works for me:

$ a=$'\n'
$ printf "%d\n" "'$a"
10
$ printf -v byte "%d" \'"$a"
$ echo "$byte"
10

Check how you're handling whatever you think should contain the newline
before converting it.

-- 
D.