>I thought it would use/fetch the value of i just before the addition would be done, ie the latest value
>
>

>int i = 1;
>i += ++i;  // 3 and not 4
>

It does but i += overload means i = i + ... thus

i += ++i; // 3 and not 4

expand to:

i = i + ++i;

There is only a new value for second parameter.
Cetin