Date:	Wed, 26 Jan 2000 14:19:36 +0200 (IST)
From:	Eli Zaretskii <eliz AT is DOT elta DOT co DOT il>
X-Sender:	eliz AT is
To:	Petr Maxa <maxa AT sse-za DOT sk>
cc:	djgpp AT delorie DOT com
Subject:	Re: signed - unsigned chars
In-Reply-To:	<001501bf6710$21cbb840$0307028a@prmivv03>
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On Tue, 25 Jan 2000, Petr Maxa wrote:

> int main()
> {
> uchar buf[256];
> 
>   code_message(buf,0x81,(uchar *)"Hello world\n",13); /* 1-st overtyping */
> 
>   if((buf[0]==0xff) && (buf[1]==0x81))  /*!! if we use signed char, this
> comparation would be useless !! */
>     printf((char*)buf+3);       /* 2-nd overtyping */
>  return 0;
> }

I was referring to the last cast, inside `printf': why did you need
it?  AFAIK, "printf(buf+3)" should work with no problems.  Doesn't it?

> 2. printf function request the "signed char *" type as the first parameter.

AFAIK, printf can do well with unsigned char as well.

> For curiosity (or off topic), there is also another problem. If You have
> signed char type variable with value 0xff (-1) and You convert it to int,
> You get 0xffffffff that is also -1, but in our case it might produce buggy
> code.

Your code mixes signed and unsigned integral types.  This always
spells trouble.  You should avoid mixing signed and unsigned types at
all costs.

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