Question

I got the following code:

int main(int argc, char *argv[])
{
    char c = 128;

    c = c >> 1;

    printf("c = %d\n", c);

    return 0;
}

Running the above code on Windows XP 32 bit, I got the result: -64. Why -64?

Answer 1

Because the char type is a signed 8-bit integer (in the implementation of C that you are using). If you try to store the value 128 in it, it will actually be -128.

The bits for that would be:

10000000

Shifting a negative number will keep the sign bit set (as your implementation uses an arithmetic shift):

11000000

The result is -64.