PORTA & 0x03 == 0x03

[sonos]

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

 

[Chris Dollin]

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).

 

[Richard Heathfield]

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.

 

[Frederick Gotham]

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 ))

 

[Thad Smith]

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) ...

 

[Fred Kleinschmidt]

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.

 

[Tom St Denis]

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]

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

 

[Giorgio Silvestri]

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

 

[Gernot Frisch]

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) ...