RE: [sv-bc] Question on 1800 section 8.3

I would say that the fact that i++ is a post-increment ensures only that
the value returned from this expression is before the increment.
However, nothing seems to say that the increment itself (the blocking
assignment) has to happen before the whole RHS expression is evaluated.
On the contrary, LRM explicitly says that the ordering of the assignment
operations relative to any other operations within an expression is
undefined, implying that the assignments may happen even before other
operations in the same expression are performed (just like it happens
with the assignment operator in your first example).

So, if i++ is evaluated first, then the value is incremented, and then
the i = i - 1 is performed, we get 10 + (i = 11 - 1) == 10 + 10 == 20. 

Still, I agree that the simple reference to the 18, 19, 20 numbers is
confusing and should probably be either removed or elaborated.

--Yulik.

-----Original Message-----
From: owner-sv-bc@server.eda.org [mailto:owner-sv-bc@server.eda.org] On
Behalf Of Stuart Sutherland
Sent: Friday, September 01, 2006 2:22 PM
To: sv-bc@server.eda.org
Subject: [sv-bc] Question on 1800 section 8.3

Section 8.3 of the 1800 standard contains the following example and
explanation:

i = 10;
j = i++ + (i = i - 1);
After execution, the value of j can be 18, 19, or 20 depending upon the
relative ordering of the increment
and the assignment statements.

Can anybody detail the three orders of execution that would lead to
these
three possible results? 

The only combinations I can come up with are:

j = 9 + 9 (i-1 is evaluated first, and used as the value of i++)

j = 10 + 9 (i used as the value for i++ first, and then i-1 is
evaluated)

It seems to me that since i++ is a post-increment, it cannot change i
until
after j has been assigned.

Thanks,

Stu
~~~~~~~~~~~~~~~~~~~~~~~~~
Stuart Sutherland
stuart@sutherland-hdl.com
+1-503-692-0898