AVR-Chat

On Jun 26, 2009, at 12:10 PM, Антощенков Роман  
Викторович wrote:

> Hello!
>
> I am coding in AVRStudio 4.16 and WinAVR-20090313
>
> Code:
>  unsigned int  i1 = 0;
>  unsigned int  i2 = 0;
>  unsigned char c1 = 0;
>
>  i1 = 1024;
>  i2 = i1 >> 2;
>  c1 = (unsigned char)i1>>2;
>
>  Why c1 = 0? Must be c1 = 256!

[...]

> 291:        c1 = (unsigned char)i1>>2;
>
> Why register R24? Must be R25

No, R24 is correct. You only ask for the least significant 8 bits of i1.

> +0000045C:   818C        LDD       R24,Y+4        Load indirect with  
> displacement
> +0000045D:   9586        LSR       R24            Logical shift right
> +0000045E:   9586        LSR       R24            Logical shift right
> +0000045F:   8389        STD       Y+1,R24        Store indirect  
> with displacement

No, the above generated code is doing exactly what you told it to. You  
cast i1 to (unsigned char) so the compiler dutifully picked up only  
the low 8 bits which were zero.

The following is not the same thing as what you wrote above, but  
perhaps what you intended?

c1 = (unsigned char)(i1>>2);

i1 equals 256. After the above c1 equals 64, but after your code c1  
equals 0.

--
David Kelly N4HHE, dkelly@HiWAAY.net
========================================================================
Whom computers would destroy, they must first drive mad.