Weird behavior of right shift operator (1 >> 32)

Question

I recently faced a strange behavior using the right-shift operator.

The following program:

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <stdint.h>

int foo(int a, int b)
{
   return a >> b;
}

int bar(uint64_t a, int b)
{
   return a >> b;
}

int main(int argc, char** argv)
{
    std::cout << "foo(1, 32): " << foo(1, 32) << std::endl;
    std::cout << "bar(1, 32): " << bar(1, 32) << std::endl;
    std::cout << "1 >> 32: " << (1 >> 32) << std::endl; //warning here
    std::cout << "(int)1 >> (int)32: " << ((int)1 >> (int)32) << std::endl; //warning here

    return EXIT_SUCCESS;
}

Outputs:

foo(1, 32): 1 // Should be 0 (but I guess I'm missing something)
bar(1, 32): 0
1 >> 32: 0
(int)1 >> (int)32: 0

What happens with the foo() function ? I understand that the only difference between what it does and the last 2 lines, is that the last two lines are evaluated at compile time. And why does it "work" if I use a 64 bits integer ?

Any lights regarding this will be greatly appreciated !

---

Surely related, here is what g++ gives:

> g++ -o test test.cpp
test.cpp: In function 'int main(int, char**)':
test.cpp:20:36: warning: right shift count >= width of type
test.cpp:21:56: warning: right shift count >= width of type

Answer 1

OK. So it's in 5.8.1:

The operands shall be of integral or enumeration type and integral promotions are performed. The type of the result is that of the promoted left operand. The behavior is undefined if the right operand is negative, or greater than or equal to the length in bits of the promoted left operand.

So you have an Undefined Behaviour(tm).