[Bug c/69828] result of shift operation has wrong type

[pinskia at gcc dot gnu.org]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=69828

Andrew Pinski <pinskia at gcc dot gnu.org> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|UNCONFIRMED                 |RESOLVED
         Resolution|---                         |INVALID

--- Comment #1 from Andrew Pinski <pinskia at gcc dot gnu.org> ---
Let's look at the last one:
   i8_t i = (argc < 0) | 0x80;
   u8_t u = i;

   i16_t t = (i16_t)u;  << Not really there but useful otherwise.
   << This is a zero extend from u8 to i16 since u8 fits in i16 so
   t = 0x80;

   i32_t x32_c = t << 8;
   t is prompted to int due to normal C promotion rules.
   so t << 8 is the same as ((int)(0x80)) << 8 or
   0x8000 aka 32768

   printf("should be negative: %7d\n",x32_c);

So GCC is correct.

>The result of a shift operation should have the type of the left operand.

No, it should go though the standard C promotion rules.  The standard promotion
rules says if the size of the type is less than int promote to int.