Question

If the value after the shift operator
is greater than the number of bits in
the left-hand operand, the result is
undefined. If the left-hand operand is
unsigned, the right shift is a logical
shift so the upper bits will be filled
with zeros. If the left-hand operand
is signed, the right shift may or may
not be a logical shift (that is, the
behavior is undefined).

Can somebody explain me what the above lines mean??

Answer 1

I’m assuming you know what it means by shifting. Lets say you’re dealing with a 8-bit chars

unsigned char c;
c >> 9;
c >> 4;
signed char c;
c >> 4;

The first shift, the compiler is free to do whatever it wants, because 9 > 8 [the number of bits in a char]. Undefined behavior means all bets are off, there is no way of knowing what will happen. The second shift is well defined. You get 0s on the left: 11111111 becomes 00001111. The third shift is, like the first, undefined.

Note that, in this third case, it doesn’t matter what the value of c is. When it refers to signed, it means the type of the variable, not whether or not the actual value is greater than zero. signed char c = 5 and signed char c = -5 are both signed, and shifting to the right is undefined behavior.