[Bug c/18065] usual arithmetic conversion not applying correctly

["schlie at comcast dot net" <gcc-bugzilla@gcc.gnu.org>]

------- Additional Comments From schlie at comcast dot net  2004-10-20 19:14 -------
Subject: Re:  usual arithmetic conversion not applying
 correctly

Please read rule 1:

 "If both operands have the same type, then no further conversion is
  needed." (no promotion to int is specified or implied as being required)

Seems plain enough to me, what makes you think that:

  char + char  => int + int ?
 short + short => int + int ?

That would imply, QI mode (and HI mode typical of shorts on 32 bit machines)
rtl instruction definitions are meaningless which isn't the case; as their
definition defines the char and short operand rtl operator -> code
transformation.

The front end has no business promoting all char and short arithmetic
operands to ints, and in fact it doesn't; so am not sure what your thinking?

-paul-

> From: pinskia at gcc dot gnu dot org <gcc-bugzilla@gcc.gnu.org>
> Reply-To: <gcc-bugzilla@gcc.gnu.org>
> Date: 20 Oct 2004 18:50:01 -0000
> To: <schlie@comcast.net>
> Subject: [Bug c/18065] usual arithmetic conversion not applying correctly
> 
> 
> ------- Additional Comments From pinskia at gcc dot gnu dot org  2004-10-20
> 18:50 -------
> From C99
> (6.5.5/3): The usual arithmetic conversions are performed on the operands.
> 
> 6.3.1.8 Usual arithmetic conversions
> Otherwise, the integer promotions are performed on both operands. Then the
> following rules are
> applied to the promoted operands: If both operands have the same type, then no
> further conversion is
> needed. Otherwise, if both operands have signed integer types or both have
> unsigned integer types, the
> operand with the type of lesser integer conversion rank is converted to the
> type of the operand with
> greater rank. Otherwise, if the operand that has unsigned integer type has
> rank greater or equal to the
> rank of the type of the other operand, then the operand with signed integer
> type is converted to the
> type of the operand with unsigned integer type. Otherwise, if the type of the
> operand with signed
> integer type can represent all of the values of the type of the operand with
> unsigned integer type, then
> the operand with unsigned integer type is converted to the type of the operand
> with signed integer
> type. Otherwise, both operands are converted to the unsigned integer type
> corresponding to the type of
> the operand with signed integer type.
> 
> This is still not wrong code as signed char signed extended to int is still
> the same number but this is
> not a target problem but a front-end problem now.
> 
> To make sure that I read the statement correctly from the standard, it says if
> the types are the same
> there is no need to promote the value at all.
> 
> -- 
>            What    |Removed                     |Added
> ----------------------------------------------------------------------------
>            Severity|minor                       |normal
>           Component|target                      |c
>             Summary|not using qi version of     |usual arithmetic conversion
>                    |divmod                      |not applying correctly
> 
> 
> http://gcc.gnu.org/bugzilla/show_bug.cgi?id=18065
> 
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-- 


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=18065