Re: Why convert to pointer type before added to a pointer

[Jeffrey A Law <law at redhat dot com>]

On Mon, 2004-08-16 at 04:03, Jie Zhang wrote:
> This simple code
> 
> void
> foo (int* a, int n)
> {
>    a[n] = 1;
> }
> 
> is transformed into
> 
> foo (a, n)
> {
>    unsigned int n.0;
>    unsigned int T.1;
>    int * T.2;
>    int * T.3;
> 
>    n.0 = (unsigned int)n;
>    T.1 = n.0 * 4;
>    T.2 = (int *)T.1;      <== Why this converting
>    T.3 = T.2 + a;
>    *T.3 = 1;
> }
> 
> after the genericizing. Thanks.
It's an unfortunate aspect of the type system we have -- basically
both operands of a binary operator need to have compatible types.

I spent some time investigating ways to fix this in the context of
solving some long standing problems with the PA, IA-64 and mn10300
ports.

The first approach is to have a separate set of operators such as
PLUS_PTR_EXPR, MINUS_PTR_EXPR which do not require their operands
to be the same type.  That's ugly simply because everywhere you 
generate or test for PLUS_EXPR or MINUS_EXPR you have to to examine
the types to determine instead you should be using PLUS_PTR_EXPR
or MINUS_PTR_EXPR.

The other approach is to relax the type system.  That's bad from the
standpoint of being able to perform internal consistency checking.

Right now we just convert the index to a pointer type and hope
everything works out in the end.  That's not particularly good
either and has led to a little hack in how we compute REG_POINTER_FLAG
when converting from trees into RTL.

jeff