--- Comment #12 from manu at gcc dot gnu dot org 2007-01-13 02:23 ---
There is no warning with GCC 4.3 and you can get a warning for the int->char
conversion by using -Wconversion. So I am going to close this. Please, feel
free to reopen if you think there is some unresolved issue.
--
--- Comment #11 from manu at gcc dot gnu dot org 2006-11-25 13:51 ---
I don't get the warning with current mainline (revision 119143).
Still, I would like to keep this bug around since it may be interesting that
Wconversion emits a warning for the int->char conversion. I would like to h
--- Additional Comments From neil at daikokuya dot co dot uk 2005-02-17
15:12 ---
Subject: Re: incorrect overflow warning
schlie at comcast dot net wrote:-
>
> --- Additional Comments From schlie at comcast dot net 2005-02-17 14:33
> ---
> (In reply to comment #8)
> > char
--- Additional Comments From schlie at comcast dot net 2005-02-17 14:33
---
(In reply to comment #8)
> char x = 0x80; warning: value changes sign during integer type conversion
Implying an analogous warning for all assignments between dissimilarly
signed variables (i.e. signed x; unsign
--- Additional Comments From neil at daikokuya dot co dot uk 2005-02-17
14:00 ---
Subject: Re: incorrect overflow warning
schlie at comcast dot net wrote:-
>
> --- Additional Comments From schlie at comcast dot net 2005-02-17 13:20
> ---
> (In reply to comment #6)
> > What
--- Additional Comments From schlie at comcast dot net 2005-02-17 13:20
---
(In reply to comment #6)
> What should get a warning is the assignment of 0x80 to a char.
Not that either, as although the two differ in sign, the value does not exceed
the type's precision.
--
http://gcc.
--- Additional Comments From neil at daikokuya dot co dot uk 2005-02-17
11:34 ---
Subject: Re: incorrect overflow warning
pinskia at gcc dot gnu dot org wrote:-
>
> --- Additional Comments From pinskia at gcc dot gnu dot org 2005-02-17
> 03:14 ---
> No the warning is correc
--- Additional Comments From igodard at pacbell dot net 2005-02-17 04:10
---
Please:
yes, the int value lwbi arising from the conversion of char(0x80) is the int
value 0xff80, i.e. int(-128); you are quite right about that. That value is
being *subtracted* from the int value upbi
--- Additional Comments From pinskia at gcc dot gnu dot org 2005-02-17
03:43 ---
static const int lwbi = lwb;
lwb = 0x80
lwbi is the signed extended version of lwb to the size of int so you will still
get 0xFF80.
So this is still not a bug.
--
What|Removed
--- Additional Comments From igodard at pacbell dot net 2005-02-17 03:37
---
WADR, but "char" on my (x86 Linux) machine in fact signed. So I tried:
#include
static const char lwb = 0x80;
static const char upb = 0x7f;
static const int lwbi = lwb;
static const int upbi = upb;
static cons
--- Additional Comments From pinskia at gcc dot gnu dot org 2005-02-17
03:14 ---
No the warning is correct as char on a lot of targets is signed by default so
promoting (char)0x80 to int
will give 0xFF80.
--
What|Removed |Added
-
--- Additional Comments From igodard at pacbell dot net 2005-02-17 03:09
---
Turns out that the promotion rules aren't the problem, because you still get the
same message even with *explicit* promotion - the code:
#include
static const char lwb = 0x80;
static const char upb = 0x7f;
st
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