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Hi,
I hope this is the right usenet group for posting this code...

IF (PORTA & 0x03 == 0x03) ...

... I want to verify if PORTA (0x00 to 0xFF) has bit 1 and bit 2 set to 1.

Is this the proper code?

Thanks

Thad Smith wrote:

> sonos wrote:
>
>> I hope this is the right usenet group for posting this code...
>>
>> IF (PORTA & 0x03 == 0x03) ...
>>
>> .. I want to verify if PORTA (0x00 to 0xFF) has bit 1 and bit 2 set to 1.
>>
>> Is this the proper code?
>
> No. "if" must be lowercase. You must parenthesize to get the operation
> you want:
>
> if ((PORTA & 0x03) == 0x03) ...

And you don't /need/ to write the number in hex:

if ((PORTA & 3) == 3) ...

although your style may prefer it (especially if there are wider bit-patterns
around).

--
Chris "seeker" Dollin
A rock is not a fact. A rock is a rock.

sonos said:

> Hi,
> I hope this is the right usenet group for posting this code...
>
> IF (PORTA & 0x03 == 0x03) ...

....looks oddly familiar. The same expression - and I mean precisely the same
expression - appeared in an OP a week or three ago.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

sonos posted:

> Hi,
> I hope this is the right usenet group for posting this code...
>
> IF (PORTA & 0x03 == 0x03) ...
>
> .. I want to verify if PORTA (0x00 to 0xFF) has bit 1 and bit 2 set to 1.
>
> Is this the proper code?


No no on!

Open up your favourite operator precedence table:

http://www.difranco.net/cop2220/op-prec.htm

You'll notice that "==" has higher precedence than "&", so you'll need
parentheses.

if(3 == (PORTA & 3 ))

--

Frederick Gotham

sonos wrote:

> I hope this is the right usenet group for posting this code...
>
> IF (PORTA & 0x03 == 0x03) ...
>
> .. I want to verify if PORTA (0x00 to 0xFF) has bit 1 and bit 2 set to 1.
>
> Is this the proper code?

No. "if" must be lowercase. You must parenthesize to get the operation
you want:

if ((PORTA & 0x03) == 0x03) ...

--
Thad

"Chris Dollin" <> wrote in message
news:e92s1q$a2q$...
> Thad Smith wrote:
>
>> sonos wrote:
>>
>>> I hope this is the right usenet group for posting this code...
>>>
>>> IF (PORTA & 0x03 == 0x03) ...
>>>
>>> .. I want to verify if PORTA (0x00 to 0xFF) has bit 1 and bit 2 set to
>>> 1.
>>>
>>> Is this the proper code?
>>
>> No. "if" must be lowercase. You must parenthesize to get the operation
>> you want:
>>
>> if ((PORTA & 0x03) == 0x03) ...
>
> And you don't /need/ to write the number in hex:
>
> if ((PORTA & 3) == 3) ...
>
> although your style may prefer it (especially if there are wider
> bit-patterns
> around).

The OP said:
I want to verify if PORTA (0x00 to 0xFF) has bit 1 and bit 2 set to 1.

One could imply from this that the correct answer would be:
if ( PORTA & 3 ) ...
because the OP did not specify that ONLY bits 1 and 2 must be set.

>
> --
> Chris "seeker" Dollin
> A rock is not a fact. A rock is a rock.
>
--
Fred L. Kleinschmidt
Boeing Associate Technical Fellow
Technical Architect, Software Reuse Project

Fred Kleinschmidt wrote:
> The OP said:
> I want to verify if PORTA (0x00 to 0xFF) has bit 1 and bit 2 set to 1.
>
> One could imply from this that the correct answer would be:
> if ( PORTA & 3 ) ...
> because the OP did not specify that ONLY bits 1 and 2 must be set.


I think you had a brain fart. a & 3 will be true if a & 1 is true
and/or if a & 2 is true.

The two cases are

if ((a & 3) == a) { ... }

which means ONLY both bits are set and no others or

if ((a & 3) == 3) { ... }

which means that the two lower bits must be set

Tom

Tom St Denis wrote:
> if ((a & 3) == a) { ... }
>
> which means ONLY both bits are set and no others or

foot planted in mouth...

....

if (a == 3) { ... }

Teach me to be all superior. hehehehehe

Tom

"Tom St Denis" <> ha scritto nel messaggio
news: ups.com...
> Tom St Denis wrote:
> > if ((a & 3) == a) { ... }
> >
> > which means ONLY both bits are set and no others or
>
> foot planted in mouth...
>
> ...
>
> if (a == 3) { ... }
>
> Teach me to be all superior. hehehehehe
>
> Tom
>

Not completely equivalent.

I presume 'a' and 'PORTA', see the OP question, are
device registers, probably declared with volatile.

Consequently

if (a == 3) {
/* ... */
}

is different from

if ((a & 3) == a) {
/* ... */
}

In second form the memory-mapped I/O port is read twice.
It can be important.


Giorgio Silvestri

"sonos" <> schrieb im Newsbeitrag
news:d5ydnXaea-...
> Hi,
> I hope this is the right usenet group for posting this code...
>
> IF (PORTA & 0x03 == 0x03) ...
>
> .. I want to verify if PORTA (0x00 to 0xFF) has bit 1 and bit 2 set
> to 1.
>
> Is this the proper code?

== comes before &, thus you want:
if ( (PORTA&0x03) == 0x03) ...