RCM2000 prgram for RCM6700 module

Hello Forum,
I am something of a newbie when it comes to programming in Dynamic C. I have downloaded a simple program that worked well for the RCM2000. I am now using a RCM6700 module and would like to know whether I can use it to compile and run the same program. I have compiled it already and recieved no error messages. However, I would like to know if it will work the same way on the RCM6700 module as it did with the RCM2000 one.
The program is below. If anyone has the time to look over it, I would be very grateful.

Regards,
Ricardo

#define THRESHOLD 10
main()
{
static int j, d;

#asm
ld a,0x51
ioi ld (WDTTR),a
ld a,0x54
ioi ld (WDTTR),a
#endasm

#nodebug
WrPortI(SPCR, &SPCRShadow, 0x84);
WrPortI(PEFR, &PEFRShadow, 0x00);
WrPortI(PEDDR, &PEDDRShadow, 0x00);
BitWrPortI(PADR, &PADRShadow, 1, 1); // Preset the gate output high
while(1) {
while(1) {
/* Bring PA0 low, then high again, leaving PA1 high. Making this inline assembly reduces the
time this takes from 15 us to 1 us. /
#asm
ld a, 0x03
ioi ld (PADR), a
ld (PADRShadow), a
ld a, 0x02
ioi ld (PADR), a
ld (PADRShadow), a
nop
nop
nop
nop
nop
nop
#endasm
d = RdPortI(PEDR);
j = 0; // delay
if(d > THRESHOLD) {
break;
}
}
/
If the count goes over the threshold for two consecutive cycles, then trip the gate.
For speed, the second test is inlined instead of being made into some sort of loop with a
variable check. */
#asm
ld a, 0x03
ioi ld (PADR), a
ld (PADRShadow), a
ld a, 0x02
ioi ld (PADR), a
ld (PADRShadow), a
nop
nop
nop
nop
nop
nop
#endasm
d = RdPortI(PEDR);
j = 0;
if(d > THRESHOLD) {
//BitWrPortI(PADR, &PADRShadow, 0, 1); // Set the gate output low
#asm
xor a, a
ioi ld (PADR), a
ld (PADRShadow), a
#endasm

		while(1) { /* check to see if the reset switch has been pressed */
			d = RdPortI(PBDR);
			if(!(d & 0x04)) {
				BitWrPortI(PADR, &PADRShadow, 1, 1);
				break;
			}
		}
	}
}

}

Hello,

You will need to make a comparison of the registers and general IO of the RCM2000 vs. the RCM6700 and make sure that they are comparable.