Why -1 >> 1 is -1 ? And 1 >> 1 is 0!

Question

I have next code:

std::cout << (-10 >> 1) << std::endl;
std::cout << (-9 >> 1) << std::endl;
std::cout << (-8 >> 1) << std::endl;
std::cout << (-7 >> 1) << std::endl;
std::cout << (-6 >> 1) << std::endl;
std::cout << (-5 >> 1) << std::endl;
std::cout << (-4 >> 1) << std::endl;
std::cout << (-3 >> 1) << std::endl;
std::cout << (-2 >> 1) << std::endl;
std::cout << (-1 >> 1) << std::endl;

The result is:

-5
-5
-4
-4
-3
-3
-2
-2
-1
-1

But why?

-1 is 1111 1111 (for 1 byte), -1 >> 1 must be : 1011 1111 and it is not -1 or 0! (sign-bit is not shifted, I know)

Can someone explain to me how this works?

Answer 1

Standard 5.8/3 (Shift operators) :

The value of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a nonnegative value, the value of the result is the integral part of the quotient of E1 divided by the quantity 2 raised to the power E2. If E1 has a signed type and a negative value, the resulting value is implementation-defined.

So to the question "why ?", the standard answer is : why not.

Answer 2

Right-shifting a negative number is implementation-defined.

Implementations which shift in a sign-extended bit to the leftmost bit, work as you reported.

As for why it is done this way, it's because right-shifting can be used to divide by a power of 2 with round-towards-negative-infinity (e.g. like floor()) rounding behavior:

(-8 >> 2) == -2
(-9 >> 2) == -3
(-10 >> 2) == -3
(-11 >> 2) == -3
(-12 >> 2) == -3

See this SO question.