Re: Bit shift compiler bug or a corner case?

[Ian Lance Taylor <iant at google dot com>]

Rohit Garg <rpg.314@gmail.com> writes:

> The following code outputs 0,1,32,33. Which is counter intuitive to
> say the least. But if I replace the literal 1 with the type annonated
> constant "ONE", the loop runs fine and outputs 0 and 1, which is what
> I would expect.
>
> This is with gcc 4.6.2 and -std=c++0x.
>
> #include<iostream>
> #include<cstdint>
> using namespace std;
> int main()
>     {
>     int64_t bitmask = 3;
>     int64_t k;
>     const int64_t ONE = 1;
>     cout<<"bitmask = "<<bitmask<<endl;
>
>     for(k=0; k<64; k++)
>         {
>         if(bitmask & (1<<k))
>             {
>             cout<<"k="<<k<<endl;
>             }
>         }
>
>     return 0;
>     }

The number 1 has type int, which on your system is probably 32 bits.  In
C, using a shift which is larger than the size of the integer is
undefined behaviour.  When a program uses undefined behaviour, the
results are often surprising and counter intuitive.

When you use ONE, with type int64_t, the value has 64 bits and all the
shifts are well defined.

Ian