64bit shift problem

Question

Why this code does not write 0 as a last element but 18446744073709551615? (compiled with g++)

#include <iostream>

using namespace std;
int main(){
    unsigned long long x = (unsigned long long) (-1);
    for(int i=0; i <= 64; i++)
        cout << i << " " << (x >> i) << endl;
    cout << (x >> 64) << endl;
    return 0;
}

Answer 1

You overflow the shift. If you've noticed, GCC even warns you:

warning: right shift count >= width of type

How come? You include 64 as a valid shift, which is an undefined behavior. counting from 0 to 64 there are 65 numbers (0 included). 0 being the first bit (much like arrays).

#include <iostream>

using namespace std;
int main(){
    unsigned long long x = (unsigned long long) (-1);
    for(int i=0; i < 64; i++)
        cout << i << " " << (x >> i) << endl;
    cout << (x >> 63) << endl;
    return 0;
}

Will produce the output you'd expect.

Answer 2

This warning from the compiler should be a hint:

"warning: right shift count >= width of type"

This results in undefined behavior:

http://sourcefrog.net/weblog/software/languages/C/bitshift.html

Answer 3

I get:

test.c:8: warning: right shift count >= width of type

so perhaps it's undefined behavior?