Why is i = v[i++] undefined?

Question

From the C++ (C++11) standard, §1.9.15 which discusses ordering of evaluation, is the following code example:

void g(int i, int* v) {
    i = v[i++]; // the beha         


        

Answer 1

In this example I would think that the subexpression i++ would be completely evaluated before the subexpression v[...] is evaluated, and that the result of evaluation of the subexpression is i (before the increment), but that the value of i is the incremented value after that subexpression has been completely evaluated.

The increment in i++ must be evaluated before indexing v and thus before assigning to i, but storing the value of that increment back to memory need not happen before. In the statement i = v[i++] there are two suboperations that modify i (i.e. will end up causing a store from a register into the variable i). The expression i++ is equivalent to x=i+1, i=x, and there is no requirement that both operations need to take place sequentially:

x = i+1;
y = v[i];
i = y;
i = x;

With that expansion, the result of i is unrelated to the value in v[i]. On a different expansion, the i = x assignment could take place before the i = y assignment, and the result would be i = v[i]