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Re: [RFC][BZ #17943] Use long for int_fast8_t
- From: Richard Earnshaw <Richard dot Earnshaw at foss dot arm dot com>
- To: OndÅej BÃlka <neleai at seznam dot cz>, Rich Felker <dalias at libc dot org>
- Cc: libc-alpha at sourceware dot org
- Date: Wed, 11 Feb 2015 18:26:01 +0000
- Subject: Re: [RFC][BZ #17943] Use long for int_fast8_t
- Authentication-results: sourceware.org; auth=none
- References: <20150208110426 dot GA28729 at domone> <20150209181324 dot GE23507 at brightrain dot aerifal dot cx> <20150211133409 dot GA24480 at domone>
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.