On 10/12/16 09:52, Luc Van Oostenryck wrote:
> A compound assignment like, for example:
> x /= a;
> should have the same effect as the operation followed by the
> assignment except that the left side should only be evaluated
> once. So the statement above (assuming 'x' free of side-effects)
> should have the same effect as:
> x = x / a;
>
> In particular, the usual conversions should applied. So, if the
> type of 'x' and 'a' is, respectively, 'unsigned int' (32 bit) and
> 'long' (64 bit), the statement above should be equivalent to:
> x = (unsigned int) ((long) x / a);
>
> But what is really done currently is something like:
> x = x / (unsigned int) a;
> In other words, the left-hand side is casted to the same type as the
> rhs and the operation is always done with this type, neglecting the
> usual conversions and thus forcing the operation to always be done
> with the rhs type, here 'unsigned int' instead of 'long'.
I have read this paragraph repeatedly, but I just can't understand
what you are saying, unless I swap left-hand-side for right-hand-side
and vice-versa. :-P
> For example, with the values:
> unsigned int x;
^^^^^^^^^^^^^^^
unsigned int x = 1;
> long a = -1;
>
> We have:
> x = 1 / (unsigned int) (-1);
> x = 1 / 0xffffffff;
> x = 0;
> instead of the expected:
> x = (unsigned int) (1L / -1L);
x = (unsigned int) ((long)1 / -1L);
> x = (unsigned int) (-1L);
> x = 0xffffffff;
>
ATB,
Ramsay Jones
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