powerful expressions

[Blair P. Houghton]

In article <462 at cpsolv.UUCP> rhg at cpsolv.uucp (Richard H. Gumpertz) writes:
>In article <5040 at buengc.BU.EDU> bph at buengc.bu.edu (Blair P. Houghton) writes:
>>>In <12855 at cit-vax.Caltech.Edu>, wen-king at cit-vax.Caltech.Edu (King Su) shares:
>>>>	((qhead) ? (qtail = qtail->next = qnew)
>>>>		 : (qtail = qhead       = qnew))->next = 0;
>...
>>The assignment operators group right-to-left.  That is
>>	qtail->next = qnew
>>is performed before
>>	qtail = qtail->next
>
>X3J11/88-158, page 10, line 22: "The grouping of an expression does not
>completely determine its evaluation."

This is distasteful and will continue to be distasteful so long as
assignment is considered an operator.  Could someone please email me a
copy of the rationale for this rule?

>X3J11/88-158, page 53, line 30: "The side effect of updating the stored
>value of the left operand [of an assignment operator] shall occur
>between the previous and the next sequence point."

Huh?

>X3J11/88-158, page 53, line 33: "The order of evaluation of the operands
>[of an assignment operator] is unspecified."

This one speaks of the left and right sides of a single
assignment-operator, and seems to controvert the (formerly) specified
right-to-left grouping of strings of assignments.

>Much as I would like to agree that the code is safe, I can't find anything
>in the standard that guarantees it.  I think the standards committee may
>have (intentionally?) blown this one; can anyone prove me wrong?

Especially elucidations on the possibility of that "intentionally."

Doug?

>The following should always work (assuming NULL is defined by the
>appropriate #include):
>   (qhead ? qtail->next = qnew : qhead = qnew), (qtail = qnew)->next = NULL;

I think it's time just to do:

	if (qhead)
	    qtail->next = qnew;
	else
	    qhead = qnew;

	qtail = qnew;
	qtail->next = NULL;

Except for the loss of the extra capabilities of an expression.

				--Blair
				  "No pain, uh, no pain."