PIC24EP256GP204, MPLABX v6.00, XC16 v2.10, PICKIT3
In the code below DutyCycle is int8_t and OC2RS, OC2R are processor registers uint16t.
OC2RS is 0xFFFF
DutyCycle is 75
Gives OC2R = Temp = 0x028F - INCORRECT
With commented out code enabled and the 'Temp = OC2RS * DutyCycle / 100' line commented out gives Temp = 0xBFFF - CORRECT.

How should I perform this calculation?
Is there any disadvantage in splitting up the calculation?

    uint32_t Temp;
   
#ifdef SWAP_DIRECTION_LEFT
#ifdef SWAP_PWM_LEFT
    //Temp = OC2RS;
    //Temp *= DutyCycle;
    //Temp /= 100;
    Temp = OC2RS * DutyCycle / 100;
    OC2R = Temp;

 

Temp = OC2RS;
guarantees that the calculation will be done using uint32_t.
Otherwise it is using uint16_t.

 

And the multiply overflows.

 

It's easy to be bit in C.


https://wiki.sei.cmu.edu/confluence/display/c/INT02-C.+Understand+integer+conversion+rules

 

PIC24EP256GP204, MPLABX v6.00, XC16 v2.10, PICKIT3
In the code below DutyCycle is int8_t and OC2RS, OC2R are processor registers uint16t.
OC2RS is 0xFFFF
DutyCycle is 75
Gives OC2R = Temp = 0x028F - INCORRECT
With commented out code enabled and the 'Temp = OC2RS * DutyCycle / 100' line commented out gives Temp = 0xBFFF - CORRECT.

How should I perform this calculation?
Is there any disadvantage in splitting up the calculation?

    uint32_t Temp;
  
#ifdef SWAP_DIRECTION_LEFT
#ifdef SWAP_PWM_LEFT
    //Temp = OC2RS;
    //Temp *= DutyCycle;
    //Temp /= 100;
    Temp = OC2RS * DutyCycle / 100;
    OC2R = Temp;

What happens when you split calculate Temp * Temp * (Temp/100)
Temp and Dutycycle are Integers
what happens when you divide integer 1264/100 is answer 12 or 13?


Picbuster

 

Use the same code but cast.

Temp = (int32_t) DutyCycle * OC2RS / 100;

Casting the variable and not the register.

 

What happens when you split calculate Temp * Temp * (Temp/100)
Temp and Dutycycle are Integers
what happens when you divide integer 1264/100 is answer 12 or 13?


Picbuster

Answer is 12.
Do the multiplication first in order to avoid rounding errors, assuming no overflows.

To do rounding, add one half of the divisor first before dividing.