Why?Originally Posted by vpshastry
Quzah.
Why?
Quzah.
Hope is the first step on the road to disappointment.
The problem is that exactly when the increment occurs is not specified. We only know that the increment occurs at some point AFTER the initial value of i is loaded on the RHS. So there are two possibilities:
The original value of i is stored in a temporary, then i is incremented, then the temporary is stored in i. The result is no change in the value whatsoever.
The original value of i is stored in a temporary, then the temporary is stored in i, then i is incremented. The result is that i is incremented.
I believe this is an example of unspecified behavior, as opposed to undefined.
Code://try //{ if (a) do { f( b); } while(1); else do { f(!b); } while(1); //}
No, I think that the behaviour is undefined. Consider: we know that between the previous and next sequence point, a value will be assigned to i, thus i will be modified. We also know that "side effect of updating the stored value of the operand shall occur between the previous and the next sequence point". Therefore, i is modified twice between consecutive sequence points, resulting in undefined behaviour.
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Well, try to come up with one example where i=i; could be useful.
It adds nothing to the increment operation on it's own, and it's behaviour is apparently undefined, so it's best avoided it seems.
