Why doesn't left bit shift << shift beyond 31 for long int datatype?

Question

I want to use the following code in my program but gcc won't allow me to left shift my 1 beyond 31.

sizeof(long int) displays 8, so doesn't that mean I can left shift till 63?

#include <iostream>

using namespace std;

int main(){
    long int x;
    x=(~0 & ~(1<<63));
    cout<<x<<endl;
    return 0;
}

The compiling outputs the following warning:

left shift `count >= width` of type [enabled by default] `x=(~0 & ~(1<<63))`;
                                                                    ^

and the output is -1. Had I left shifted 31 bits I get 2147483647 as expected of int.

I am expecting all bits except the MSB to be turned on thus displaying the maximum value the datatype can hold.

Answer 1

Although your x is of type long int, the 1 is not. 1 is an int, so 1<<63 is indeed undefined.

Try (static_cast<long int>(1) << 63), or 1L << 63 as suggested by Wojtek.

Answer 2

You can't use 1 (int by default) to shift it beyond the int boundaries.

There's an easier way to get the "all bits except the MSB turned on" for a specific datatype

#include <iostream>
#include <limits>

using namespace std;

int main(){
    unsigned long int max = std::numeric_limits<unsigned long int>::max();
    unsigned long int max_without_MSB = max >> 1;
    cout<< max_without_MSB <<endl;
    return 0;
}

note the unsigned type. Without numeric_limits:

#include <iostream>
using namespace std;

int main() {

    long int max = -1;
    unsigned long int max_without_MSB = ((unsigned long int)max) >> 1;
    cout << max_without_MSB << endl;

    return 0;
}