Is it possible to combine two 8 bit ports in 877A as a single 16 bit port?

[vinodstanur]

Is it possible to combine two 8 bit ports in 877A as a single 16 bit port so that i could perform 16bit shift and other bitwise operations...

 

[betwixt]

Not normally but you can of course write software that 'carries' the end bit from one port into the other.
To output a 16 bit number to two ports (using PORT B and C in my example) you can do this:
PORTB = (Number & 0xFF00) >> 8;
PORTC = Number & 0xFF;

To combine them into a 16 bit number, do the opposite:

Number = (PORTB << 8 ) | PORTC ;

Brian.

 

[millwood]

yes. just create a shadow 16-bit variable and perform right or left shift on it.

using the same example, you can actual great a 32-bit or even a 64-bit variable and shift it left/right.

---------- Post added at 21:05 ---------- Previous post was at 20:31 ----------

a quick example would be:

#define MSB_PORT PORTC
#define LSB_PORT PORTB
#define OUT_WORD(word_t) {MSB_PORT = 0xff00 & (word_t); LSB_PORT = 0x00ff & (word_t);}
...

unsigned char sPort_16bit=0;

...

  sPort_16bit = sPort_16bit << 1;  //left shift shadow variable
  OUT_WORD(sPort_16bit);          //output the shadow variable on port
...

so all you need is to update the shadow variable.

---------- Post added at 21:06 ---------- Previous post was at 21:05 ----------

and you can scale up the approach to any bit length fairly easily.

 

[yager]

Is it possible to combine two 8 bit ports in 877A as a single 16 bit port so that i could perform 16bit shift and other bitwise operations...

This is a dirty way which it depends on MCU,compiler,endian.....
Poor portability, good performance....
EX: to use HTC for PIC16

volatile unsigned short int PORTCB @ 0x00D; // PORTCB.L(PORTB):0x0D, PORTCB.H(PORTC):0x0E

void test( void )
{
   PORTCB <<= 1;
}

 

[incomplete_jerin]

Just initialize a 16 bit variable & out the lower 8 bit on a port by 'ANDing' with 0x00ff & upper 8 bit on another port by 'ANDing' with 0xff00.
Regards,
Jerin.