Bitshift operation with signed and unsigned

Question

I was doing a bitshift operation of an int and was surprised that it didn't come out as expected.

int i, res;
i = 0x80000000;
res = i>>1;                //results in 0xc0000000
res = (unsigned int) i>>1; //results in 0x40000000

how is it possible that a shift of a bit in an integer does only work to the 31st bit?

Answer 1

What you are seeing is probably arithmetic bit shift .

when shifting to the right, the leftmost bit (usually the sign bit in signed integer representations) is replicated to fill in all the vacant positions (this is a kind of sign extension).

The C99 standard 6.5.7§5 says:

The result of E1 >> E2 is E1 right-shifted E2 bit positions. [...] If E1 has a signed type and a negative value, the resulting value is implementation-defined.

So the result could be anything the compiler writers decided it to be. They probably decided to extend the sign bit, the compiler doc should mention it.