TITLE: Sequence points and expression side effects

PROBLEM: andrzej@cs.UAlberta.CA (Semeniuk Andrzej)

int j = 0;
(++j)++;
cout << "j = " << j << "\n";

and obtained the following output:
j = 3
I expected a 2.


RESPONSE: mayoff@sylvester.cc.utexas.edu (rob)

I suggest that the parenthesis guarantee that the inner expression must
be executed before its containing expression.  You can't compute (a+b)+c
without first computing (a+b), right?  So to evaluate (++j)++, you must
first compute (++j), which includes executing the side effect of storing
the new value of j.


RESPONSE: jimad@microsoft.com (Jim Adcock), Microsoft Corporation, 9 Jul 93

No, this last sentence is in error.  ARM states that the value of (++j)
is an lvalue equal to (j+1).  ARM doesn't state that the lvalue is
a reference to j.  Thus while (++j) must return a value (j+1), the
point in time that j itself is incremented remains undefined
[excepting that it must happen between sequence points]

IE the compiler could implement this preincrement part as if:

	j = j+1;
	return j;
	implement other parts of the expression

or it could implement it as if:

	return (temp = j+1);
	implement other parts of the expression
	j = temp;

or the compiler could do whatever else it might choose, since the
overall expression is an unconstrained error [my claim from the 
ANSI C base doc]

Note that there *are* some expressions where multiple side effects
are allowed, such as ARM alludes to.  The following expression
is such an example:

	j++, j++, j++;

Since each comma in the expression represents a sequence point,
the side effect must be completed by each comma, and thus this
expression is legal and well-defined. Legal both by ARM and
X3.159-1989.