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Re: Is this a compiler bug?
- From: Steve Kargl <sgk at troutmask dot apl dot washington dot edu>
- To: Andrew Pinski <pinskia at gmail dot com>
- Cc: GCC Mailing List <gcc at gcc dot gnu dot org>
- Date: Sun, 21 Sep 2014 18:56:27 -0700
- Subject: Re: Is this a compiler bug?
- Authentication-results: sourceware.org; auth=none
- References: <20140922012346 dot GA4376 at troutmask dot apl dot washington dot edu> <CA+=Sn1nT8by8dUHugr_Ye8TYmSbGS6Axbbmau7d+-d7WCUVPfQ at mail dot gmail dot com>
+ is a binary operator. 0x3ffe is a hexidecimal-constant according
to 6.6.4.1 in n1256.pdf. 63 is, of course, a decimal-constant.
--
steve
On Sun, Sep 21, 2014 at 06:49:54PM -0700, Andrew Pinski wrote:
> On Sun, Sep 21, 2014 at 6:23 PM, Steve Kargl
> <sgk@troutmask.apl.washington.edu> wrote:
>
> No e+ is exponent marker.
>
>
> > #include <stdio.h>
> > #include <stdint.h>
> >
> > int
> > main(void)
> > {
> > uint16_t i;
> > i = 0x3ff0+63; printf("%x\n", i);
> > i = 0x3ff1+63; printf("%x\n", i);
> > i = 0x3ff2+63; printf("%x\n", i);
> > i = 0x3ff3+63; printf("%x\n", i);
> > i = 0x3ff4+63; printf("%x\n", i);
> > i = 0x3ff4+63; printf("%x\n", i);
> > i = 0x3ff6+63; printf("%x\n", i);
> > i = 0x3ff7+63; printf("%x\n", i);
> > i = 0x3ff8+63; printf("%x\n", i);
> > i = 0x3ff9+63; printf("%x\n", i);
> > i = 0x3ffa+63; printf("%x\n", i);
> > i = 0x3ffb+63; printf("%x\n", i);
> > i = 0x3ffc+63; printf("%x\n", i);
> > i = 0x3ffd+63; printf("%x\n", i);
> > i = 0x3ffe+63; printf("%x\n", i);
> > i = 0x3fff+63; printf("%x\n", i);
> > return 0;
> > }
> >
> > --
> > Steve
--
Steve