RE: [avr-gcc-list] Re: Why is gcc promoting on an unsigned char?

[Dave Hansen]

From: "J.C. Wren" <address@hidden>
[...]
> OK, part of that was my fault, I had "i << 1" instead of "1 << i".  But
> nonetheless, we still see some extra code at 0x94, 0x9c, 0xa0, and 0xa4.  
> The
> compiler should be able to foresee the type of 'data', and not promote the
> intermediate results, right?

As others have mentioned, this is a case where the generated code is 
correct, though the compiler might be able to optimize better.  You and I 
are both aware of a compiler that, given

  unsigned char shift = 12;
  unsigned int i;
  unsigned int j;

  i = 1 << shift;
  j = 1U << shift;

will (depending on the options set) set i to 0 and j to 4096.  Gcc won't 
make that mistake.

Regards,
  -=Dave

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