🚀 go-pugleaf

In a simple expression:

	a = b = c;

where:
	int a,c;
	volatile int b;

In determining the value to be assigned to a, the standard says in
section 3.3.16, "An assignment expression has the value of the left
operand after the assignment".  Suppose 'b' is volatile, then we can't
expect the value of b after the assignment to have any relation to the
value stored into it (namely c).  So the question is, can a=b=c be
implemented as b=c, a=c.  Or does the standard force it to be
implemented as: b=c, a=b.

I give an example, suppose 'b' is an I/O port, writing to b outputs a
character, and reading b reads an input character.  This is a case
where reading the value of b may not necessarily be the value written
to it.  So the programmer writes:
	a=b=c;
expecting to output the value c, and inputting an unknown value
into a.  The value of the left hand side (of b=c), after the
assignment, may not be c.  Can an implementation "expect" the
value of "b=c" to be "c"?

Another related question, does the left hand side have to be fully
evaluated to determine the value of an assignment expression:

	a = *(complicated_expression_with_side_effects) = c;

In other words, will the complicated_expression_with_side_effects be
evaluated once or twice when the above statement is executed.
Example:

	int a,*b,c;

	a = *(b++) = c; /* will b be incremented twice? */

-- 
        "What is a DJ if he can't scratch?"  - Uncle Jamms Army
     David Lai (lai@mips.com || {ames,prls,pyramid,decwrl}!mips!lai)