Question

Suppose I have

struct A
{
    signed char a:1;
    unsigned char b:1;
};

If I have

A two, three;
two.a = 2; two.b = 2;
three.a = 3; three.b = 3;

two will contain 0s in its fields, while three will contain 1s. So, this makes me think, that assigning a number to a single-bit-field gets the least significant bit (2 is 10 in binary and 3 is 11).

So, my question is – is this correct and cross-platform? Or it depends on the machine, on the compiler, etc. Does the standard says anything about this, or it’s completely implementation defined?

_{Note: The same result may be achieved by assigning 0 and 1, instead of 2 and 3 respectively. I used 2 and 3 just for illustrating my question, I wouldn’t use it in a real-world situation}

_{P.S. And, yes, I’m interesting in both – C and C++, please don’t tell me they are different languages, because I know this 🙂}

Answer 1

The rules in this case are no different than in case of full-width arithmetic. Bit-fields behave the same way as the corresponding full-size types, except that their width is limited by the value you specified in the bit-field declaration (6.7.2.1/9 in C99).

Assigning an overflowing value to a signed bit-field leads to implementation-defined behavior, which means that behavior you observe with bit-field a is generally not portable.

Assigning an overflowing value to an unsigned bit-field uses the rules of modulo arithmetic, meaning that the value is taken modulo 2^N, where N is the width of the bit-field. This means, for example, that assigning even numbers to your bit-field b will always produce value 0, while assigning odd numbers to such bit-field will always produce 1.