[Voted into WP at March 2004 meeting.]

Proposed resolution (April 2003):

Change the example in Clause 7 [expr], paragraph 4 from

[Example:

i = v[i++];                     //  the behavior is unspecified
i = 7, i++, i++;                //   i  becomes  9

i = ++i + 1;                    //  the behavior is unspecified
i = i + 1;                      //  the value of  i  is incremented

--- end example]

to (changing "unspecified" to "undefined" twice)

[Example:

i = v[i++];                     //  the behavior is undefined
i = 7, i++, i++;                //   i  becomes  9

i = ++i + 1;                    //  the behavior is undefined
i = i + 1;                      //  the value of  i  is incremented

--- end example]

Notes from October 2002 meeting:

We should find out what C99 says and do the same thing.

I have found what looks like a bug in Clause 7 [expr], paragraph 4:

Between the previous and next sequence point a scalar object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be accessed only to determine the value to be stored. The requirements of this paragraph shall be met for each allowable ordering of the subexpressions of a full expression; otherwise the behavior is undefined. Example:

        i = v[i++];                     // the behavior is unspecified
        i = 7, i++, i++;                // i becomes 9

        i = ++i + 1;                    // the behavior is unspecified
        i = i + 1;                      // the value of i is incremented

--end example]

So which is it, unspecified or undefined?