------- 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
