Question

In am working on something and came across code similar to the following:

#define MODULUS(a,b)        ((a) >= 0 ? (a)%(b) : (b)-(-a)%(b))

unsigned char w;
unsigned char x;
unsigned char y;
char z;

/* Code that assigns values to w,x and y.  All values assigned 
   could have been represented by a signed char. */

z = MODULUS((x - y), w);

It is my understanding that the arithmetic (x - y) will be accomplished prior to any type conversion and the macro will always evaluate to (a)%(b) — as the result will be an unsigned char which is always greater than or equal to zero. However, the code functions as intended and I think that my understanding is flawed. So…

My questions are these:

1. Does an implicit type conversion to signed char occur before the expression is evaluated?

2. Is there a situation (for example, if the unsigned values were large enough that they could not be represented by a signed value) where the above code would not work?

Answer 1

Does an implicit type conversion to signed char occur before the expression is evaluated?

No, a conversion to int occurs before the expression x - y is evaluated¹. Thus the result can be negative.

Is there a situation (for example, if the unsigned values were large enough that they could not be represented by a signed value) where the above code would not work?

If sizeof int == 1, the integer promotion would promote the unsigned chars to unsigned ints, and that could produce wrong results because before the modulus by w, a modulus by UINT_MAX + 1 is performed due to unsigned arithmetic.

¹ The default integer promotion.