Question

When I multiply two unsigned chars in C like this:

unsigned char a = 200;
unsigned char b = 200;
unsigned char c = a * b;

Then I know I will have an overflow, and I get (40’000 modulo 256) as a result. When I do this:

unsigned char a = 200;
unsigned char b = 200;
unsigned int c = (int)a * (int)b;

I will get the correct result 40’000. However, I do not know what happens with this:

unsigned char a = 200;
unsigned char b = 200;
unsigned int c = a * b;

Can I be sure the right thing happens? Is this compiler dependent? Similarly, I don’t know what happens when doing a subtraction:

unsigned char a = 1;
unsigned char b = 2;
int c = a - b;

When making “c” an unsigned char, I will probably get 255 as a result. What happens when I use an int like this?

Answer 1

Argument of arithmetic operators get the “usual arithmetic promotions”.

In cases where int can represent all the values of all the operands (at it is the case for your example in most implementations), arguments are first converted to int. So in both cases, you get the correct result.