Boehm, Hans wrote:
>> In fact,
>>
>> if( y != 0 )
>> {
>> x += y;
>> }
>>
>> can be transformed to x += y (at source level) under the alternate
>> interpretation, but not under the original.
> I don't understand this comment.
>
> Assume a second thread does ++x; and y is always zero.
You are right, the transformation is not allowed. I was mistaken.