Submission Date: 10 Dec 92
Submittor: WG14
Source: X3J11/91-009 (Randall Meyers)
Question 1
Subclause 6.3, page 38, lines 18-27 state some very important rules governing
how a strictly conforming program can access the value of an object. The
basic theme of the rules is that an object's value may only be accessed
through an lvalue of the appropriate type. These rules are required to
permit C programs to be optimized.
The rules depend on the ``declared type of the object.'' This seems to
make the rules not apply if the object was not declared, which is the case
for an object allocated using malloc().
Do the rules somehow apply to dynamically allocated objects? Is a compiler
free to optimize the following function:
void f(int *x, double *y)
{
*x = 0;
*y = 3.14;
*x = *x + 2;
}
into the equivalent function:
void f(int *x, double *y)
{
*x = 0;
*y = 3.14;
*x = 2; /* *x known to be zero */
}
Or must an optimizer prove that pointers are not pointing at dynamically
allocated storage before performing such optimizations?
Response
Case 1: unions f(&u.i, &u.d)
Subclause 6.3.2.3, page 42, lines 5-6:
... if a member of a union object is accessed after a value has been stored in a different member of the object, the behavior is implementation-defined.
Therefore, an alias is not permitted and the optimization is allowed.
Case 2: declared objects f((int *)&d, &d)
Subclause 6.3, page 38, lines 18-27 list specific ways in which declared
objects can be accessed. Therefore, an alias is not permitted and the optimization
is allowed.
Case 3: any other, including malloced objects f((int
*)dp, dp)
We must take recourse to intent. The intent is clear from the above two
citations and from Footnote 36 on page 38:
The intent of this list is to specify those circumstances in which an object
may or may not be aliased.
Therefore, this alias is not permitted and the optimization is allowed.
In summary, yes, the rules do apply to dynamically allocated objects.