Jan-Erik,
     I think I finally understand it. In the ISR I only need the=20
instruction count++. It will be updated automatically every time the=20
timer overflows. Either before I call the delay function I need to reset=20
the timer or I need to reset the timer inside the delay function just=20
before I enter the loop. I think resetting the timer in the delay=20
function would be a better choice as I only need to put these=20
instructions in one time. Otherwise I would need to put the instructions=20
in as many times as I call the delay routine. So I would do it this way.
//Global variables
int count =3D 0;

void main(void)
{

     PORTA =3D 0;             //    Turn porta off
     TRISA =3D 255;          //    Set Porta as all inputs
     PORTC =3D 2;            //    Set RC1 is high, all other ports are low
     TRISC =3D 193;         //    RC1 -RC5 are outputs
     CMCON0 =3D 7;        //  Turn off Comparators
     ANSEL =3D 16;         //  Set RC0/AN4 as analog

// Define commands
     int    DP =3D 1 * 2;
     int Reset =3D 32 + DP;
     int GateOn =3D 16 + DP;
     int GateOff =3D DP;

while (1 =3D=3D 1)            //    Loop forever
     {
         PORTC =3D Reset;         //    Clear the Display
         PORTC =3D GateOn;      //    Start counting
         delay(1);                       // Count the frequency for one=20
second =3D HZ
         PORTC =3D GateOff;
         delay(15);                     // Display the frequency for 15=20
seconds
     }
}    //  End NewControlCode.c

void delay(int time)
{
     resettimer();
     count =3D 0;
     while(count <=3D time * 10)
     {
     }
    return;
}

void resettimer(void)
{

    TMR1H =3D 0x0B;
    TMR1L =3D 0xDC;

}

I believe that this is the way to do it. What do you think? I'm sorry=20
about confusing you. What I did was use the varuable time in the actual=20
code but I used numberseconds in the peusdocode. Again sorry about that.=20
In C you can have an empty for(); statement if you want to. This will=20
compile and run. I believe that it takes 3 instruction cycles. You can=20
use this to set up a delay loop instead of using NOPs.
Thanks,
rich!

On 7/21/2014 1:59 AM, Jan-Erik Soderholm wrote:
>
> Richard R. Pope wrote 2014-07-21 07:15:
>> Jan-Erik,
>>        I am using a function that I pass a value to that represents the
>> number of seconds that I want the delay to last. I multiply this number
>> by ten in the for loop since there are 10 100ms interrupts for a second.
>> So if I pass 1 to the delay function, count will be tested for the
>> number 10. This is one second. If I call delay again and this time I
>> pass 15 to it, count will be tested for the number 150 and this is 15
>> seconds.
>

--=20
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