Re: [RFC][BZ #17943] Use long for int_fast8_t

[Rich Felker <dalias at libc dot org>]

On Wed, Feb 11, 2015 at 06:26:01PM +0000, Richard Earnshaw wrote:
> On 11/02/15 13:34, OndÅej BÃlka wrote:
> > On Mon, Feb 09, 2015 at 01:13:24PM -0500, Rich Felker wrote:
> >> On Sun, Feb 08, 2015 at 12:04:26PM +0100, OndÅej BÃlka wrote:
> >>> Hi, as in bugzilla entry what is rationale of using char as int_fast8_t?
> >>>
> >>> It is definitely slower with division, following code is 25% slower on
> >>> haswell with char than when you use long.
> >>
> >> This claim is nonsense. It's a compiler bug. If the 8-bit divide
> >> instruction is slow, then the compiler should use 32-bit or 64-bit
> >> divide instructions to divide 8-bit types. (Note: there's actually no
> >> such thing as a division of 8-byte types; formally, they're promoted
> >> to int, so it's the compiler being stupid if it generates a slow 8-bit
> >> divide instruction for operands that are formally int!) There's no
> >> reason to use a different type for the _storage_.
> >>
> > That is also nonsense, you cannot get same speed as 32bit instruction
> > without having 8bit instruction with same performance.
> > 
> > Compiler must add extra truncation instructions to get correct result
> > which slows it down, otherwise it gets wrong result for cases like (128+128)%3
> > 
> 
> Only if the intermediate result (128+128) is assigned directly to a
> variable with less precision than int.  Otherwise the whole expression
> is calculated with int precision.

I read it as an example with small numbers that's not directly
applicable with the C type ranges. If we're actually talking about the
value 128 in C then of course 128+128 is not subject to truncation.

Rich