Struct assignment, possible DMD bug?

[Timon Gehr]

On 09/30/2012 12:51 AM, Ali Çehreli wrote:
> On 09/29/2012 11:16 AM, Timon Gehr wrote:
>  > On 09/29/2012 06:26 PM, Maxim Fomin wrote:
>
>  >>> S s = S(1,2,3);
>  >>> s = S(4, s.a, 6);
>  >>>
>  >>> assert(a == [4,1,6]);
>  >>> assert(s == S(4,4,6));
>  >>> }
>  >>>
>  >>> Setting the struct writes s.a before evaluating it while the reverse
>  >>> is true of the array assignment. Using DMD 2.0.60. GDC does what I'd
>  >>> expect and gives both as 4,1,6.
>  >>
>  >> I think this is notorious "i = ++i + ++i".
>  >
>  > There is only one mutating sub-expression.
>
> But that mutation is happening to an object that is also being read
> inside the same expression.
>
> This is one of the definitions of undefined behavior, not a compiler bug.
>
>  >> Statement s = S(4, s.a, 6) writes to s object and simultaneously reads
>  >> it.
>  >> http://dlang.org/expression.html states that assign expression is
>  >> evaluated in implementation defined-manner and it is an error to depend
>  >> on things like this.
>  >
>  > No evaluation order of the assignment expression can possibly lead to
>  > this result. This seems to be a DMD bug.
>
> The compiler seems to be applying an optimization, which it is entitled
> to as long as the language definition is not violated.
>

Technically there is no language definition to violate.

> If we are using the same object both to read and write in the same
> expression, then we should expect the consequences.
>

No. Why?

> Disclaimer: I assume that D's rules are the same as C and C++ here.
>

C and C++ do not have struct literals and if I am not mistaken,
constructor invocation is a sequence point.

Besides, this does not make any sense, what is the relevant part of the 
standard?

int c = 0;
c = c+1; // c is both read and written to in the same expression