Strange port behavior (PIC24F)

Question

I was just trying to get port IO working with this code:

#include "p24Fxxxx.h" // This header will choose the right device header

#define FCY 16000000UL // Running at 16 MIPS = Fosc/2

#include <PIC24F_plib.h>
#include <libpic30.h>

// Configuration setup
_CONFIG1( FWDTEN_OFF & GWRP_OFF & GCP_OFF & JTAGEN_OFF & ICS_PGx3 )
_CONFIG2( FCKSM_CSDCMD & OSCIOFNC_ON & POSCMOD_HS & FNOSC_FRCPLL & I2C1SEL_PRI & PLL96MHZ_ON & PLLDIV_NODIV & IESO_OFF & IOL1WAY_OFF );
_CONFIG3( SOSCSEL_IO );

int main() {
    TRISB = 0;
    PORTB = 0;

    PORTBbits.RB0 = 1;
    PORTBbits.RB4 = 1;

    while(1) {
        __delay_ms( 200 );
    }
}

So I breakpoint on the __delay_ms instruction, and take a peek at the LATB registers:

(As pictured, only LATB4 is switched on - confirmed with a multimeter)

Also, if I comment out the PORTBbits.RB4 = 1; line, LATB0 is turned on (but not LATB4)

Is the second call overwriting it somehow? Maybe, because PORTB = 0b10001; alone works.

I'm using a PIC24FJ64GB002, MPLAB X, C30 and a PICkit 3. I realise that MPLAB X isn't 100% stable - but something this simple should work.

If some PIC guru could point me in the right direction, that'd be awesome.

Answer 1

Instead of PORTxbits.RBx rather use LATxbits.LATx, when you set/reset your pins to avoid this problem.

Answer 2

Writing to an element of a bitfield will generally cause a read-modify-write operation to be performed. In particular, the line

PORTBbits.RB4 = 1;

will have the same behavior as

PORTB = PORTB | (1<<4);

My PIC is a bit rusty, but I believe that reading PORTB will return the current state of the pins, rather than the value last written?