🚀 go-pugleaf

In article <1992Sep29.222521.9221@kai.com>, david@kai.com (David Nelson) writes:
> We all know that "x = y[++i] + z[++i];" is asking for trouble. But what
> about: "a = b[i+=1,i] + c[i+=1,i];". The comma operators have introduced
> sequence points.
> 
> From ANSI 3.3 Expressions
>      Between the previous and next sequence point an object shall have its
>  stored value modified at most once by the evaluation of an expression. 
>  ...
> My interpretation would be that the sequence point only affects the
> subexpression of the operation that introduced the sequence point. Anyone
> care to confirm or reject this interpretation.

	IMHO, your interpretation is correct, in the sense, each sequence
	point introduced by the comma operator only affects that subexpression.
	In the second example, scope (if I may call it) of each comma operator
	is limited to its subexpression, that is, the expression in the array
	index.  Hence, basically, the second example boils down to be the same
	as the first example after evaluating individual comma-scoped sequence
	points.  And since, the ANSI rule stated above prohibits the case of
	a value of an object being modified twice between two sequence points,
	the second example breaks the rule causing undefined results and thus
	illegal (and shall be incarcerated promptly :-8).

	Out of curiosity, I tried the following code on our compiler, and
	as expected, the executable generated undefined results.

#include <stdio.h>
main(void)
{
   int i = 0, c;
   int a[3] = {1,2,3};

   c = a[i++,i] + a[i++,i];
   (void) printf("c = %d;i = %d\n",c,i);

   return 0;
}

	Its output was 			'c = 6;i = 2'
	After inserting a sequence point between the two operands of '+',
	the result was as expected:	'c = 5;i = 2'

	Hopefully, this proves the point.

Rest my case,
-- 
*******************************************************************************
Ravi K Mandava		email:	piobe!ravikm@ibmpa.awdpa.ibm.com
*******************************************************************************