Message-Id:	<m0ykAzO-000S41C@inti.gov.ar>
Comments:	Authenticated sender is <salvador AT natacha DOT inti DOT gov DOT ar>
From:	"Salvador Eduardo Tropea (SET)" <salvador AT inti DOT gov DOT ar>
Organization:	INTI
To:	xavicardona AT rocketmail DOT com, djgpp AT delorie DOT com
Date:	Thu, 11 Jun 1998 14:22:18 +0000
MIME-Version:	1.0
Subject:	Re: z80 to 386
In-reply-to:	<357D73EA.73E0@rocketmail.com>

Xavi Cardona <xavicardona AT rocketmail DOT com> wrote:

> I'm working on a Z80 to 80386 machine code translator.
> I make the following routine for putting the "movb $0x1,i" in the
>  string variable, but it doesn't work. Gcc says "parse error before
> asm".
> As you see, I don't want to execute the movb instruccion. I just want to 
> store the machine code of this instruccion on an array.
> main()
> {int i=5;
> char string[30]; /*more space than we really need*/
> string=asm ("movb $0x1,%0"
>         :"=g" (i)
>         :"g" (i)     );
> }
> There is any easy :) way for doing this?
> Thanks in advanced.

You can't do it! What do you really want to do? Convert a Z80 assembler source 
code into a 386 assembler source code? or binary to binary?

For the first just use normal strings (  char *string="move $0x1,?";  ) and 
replace the ? with the destination register.

For the second you must do much more work because the destination register is 
encoded in the bits.

SET
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