Does 'a[i] = i;' always result in well defined behaviour?

Question

There are several interesting questions raised here regarding undefined behaviour in C. One of them is (slightly modified)

Does the following piece of code r

Answer 1

Let us decompose the expression a[i] = i + 1 will you ?

= -- [] -- a
  \     \_ i
   \
    \_ + -- i
         \_ 1

Effectively, a[i] refers to &i however note that neither a[i] nor i+1 modifies i. i is only modified when = (the assignment itself) is executed.

Since the operands of any function need be evaluated before this function takes effect, this is actually equivalent to:

void assign(int& address, int value) { address = value; }

assign(a[i], i + 1);

It is true that = is somewhat special in that it is built-in and does not result in a function call, still the evaluation of both operands are sequenced before the actual assignment, so they are first evaluated prior to i being modified, and a[i] (which points to i location) is being assigned to.