[LLVMdev] c const

[Christopher Lamb]

On Aug 22, 2007, at 1:19 PM, Duncan Sands wrote:

> Hi Christopher,
>
>>>> In C/C++ a restrict pointer is also assumed not to alias any other
>>>> parameters, global values or local value. However the compiler must
>>>> not assume that pointers "based on" a restrict pointer do not  
>>>> alias.
>>>
>>> does "based on" mean something like: copies of the pointer made  
>>> in the
>>> function body?
>>
>> To be precise (brackets mine):
>>
>> In what follows, a pointer expression E is said to be based on object
>> P [a pointer] if (at some
>> sequence point in the execution of B [the enclosing block] prior to
>> the evaluation of E) modifying P to point to
>> a copy of the array object into which it formerly pointed would
>> change the value of E.
>> Note that ‘‘based’’ is defined only for expressions with pointer  
>> types.
>>
>> This is from http://www.open-std.org/jtc1/sc22/wg14/www/docs/
>> n1124.pdf Sec 6.7.3.1 para 3
>
> suppose A and B are array pointers, and are in fact equal.  Consider
> the code sequences
>
> (i)
> x = A[0]	(a)
> B[0] = 1	(b)
>
> and
>
> (ii)
> B[0] = 1	(b)
> x = A[0]	(a)
>
> If I understand right, (i)(b) is not based on A, either because it
> has no value (not sure what "the value of E" means) or because  
> changing
> A to point to a copy would make no difference because (i)(a) is only a
> read.  Also (i)(a) is not based on B because this is the first  
> instruction.
> On the other hand (ii)(a) is based on B because the value of x
> is 1, but would be something else if B was changed to point to a copy.
>
> Does this mean that if B is a restricted pointer, then the lines  
> can be
> swapped in (i).  That if A is a restricted pointer then the lines  
> can be
> swapped in (i) and (ii)?
>
> This sounds very similar to the Ada setup, though somehow "dual" to  
> it.

If A and B are function arguments then there is no "based on"  
relationship between pointer expressions A+0 and B+0. This is because  
changing one of the pointers, A for example, to point to a copy of  
the object it points to would change the value of the pointer  
expression A+0, but not the expression B+0.

This means that if either A or B is a restrict pointer then the  
expressions in both (i) and (ii) may be executed in any order.

This also means that if the function were called with A == B and  
either A or B is defined as restrict then the result would be  
undefined behavior.
--
Christopher Lamb



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