Date: Fri, 30 Mar 2001 01:57:32 +0200
From: Erwin Harte
Subject: Re: [DGD]Adding character to string

On Fri, Mar 30, 2001 at 01:30:43AM +0200, Frantisek Fuka wrote:
> I have this problem with DGD 1.2. The solution is probably very simple 
> but I couldn't come up with anything elegant and it drives me mad.
> 
> I need to be able to get the character according to its ASCII number. 
> For example, if I have string str = "foo" and integer num = 65, i need 
> to somehow do (str + (char)num) and get "fooA", but I keep getting 
> "foo65" and I can find no function to convert 65 to "A". I know it can 
> be done by doing
> str = "foo?"; str [3] = num; /* replaces '?' with 'A' */

Right, so what about:

    string
    chr(int c)
    {
	string s;

	s = " ";
	s[0] = c;
	return s;
    }

I agree, a bit of overhead, but once you've got that, it's very
elegant to be doing

    str += chr(65);

and have it work as intended.

> but that seems cumbersome to me. Specifically, I need to use the 
> following function for translating all output from one character coding 
> to other so it has to be quick and elegant (it also adds '\x0D' before 
> each "\n" because i use binary port for output):
> 
> string out;
> out = "";
> for (i=0; i<strlen(msg); i++) {
> if (msg[i] == '\n') out += '\x0D';
> out += (msg[i] >= 128) ? charset_output[msg[i]-128] : msg[i];
> }
> 
> Unfortunately, instead of character (for example 'A'), the string "65" 
> is being added to 'out' variable.

The reason for this is that the 'x' constants are really int values,
so what the compiler thinks you're doing is:

    <string> = <string> + <int>;

Which means you get:

    <string> = <string> + (string)<int>

So the compiler is doing what you're telling it to, not what you want
it to. ;-)

I'd do it this way:

    int    i, len;
    string out;

    out = implode(explode("\n" + msg + "\n", "\n"), "\r\n");
    for (i = 0, len = strlen(out); i < len; i++) {
        out[i] &= 0x7f;
    }

Hope that helps,

Erwin.
-- 
Erwin Harte      : `Don't mind Erwin, he gets crabby. :)'