For i = 0, why is (i += i++) equal to 0?

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This:

int i = 0;
i += i++

Can be seen as you doing (the following is a gross oversimplification):

int i = 0;
i = i + i; // i=0 because the ++ is a postfix operator and hasn't been executed
i + 1; // Note that you are discarding the calculation result

What actually happens is more involved than that – take a look at MSDN, 7.5.9 Postfix increment and decrement operators:

The run-time processing of a postfix increment or decrement operation of the form x++ or x– consists of the following steps:

  • If x is classified as a variable:

    • x is evaluated to produce the variable.
    • The value of x is saved.
    • The selected operator is invoked with the saved value of x as its argument.
    • The value returned by the operator is stored in the location given by the evaluation of x.
    • The saved value of x becomes the result of the operation.

Note that due to order of precedence, the postfix ++ occurs before +=, but the result ends up being unused (as the previous value of i is used).

A more thorough decomposition of i += i++ to the parts it is made of requires one to know that both += and ++ are not atomic (that is, neither one is a single operation), even if they look like they are. The way these are implemented involve temporary variables, copies of i before the operations take place – one for each operation. (I will use the names iAdd and iAssign for the temporary variables used for ++ and += respectively).

So, a closer approximation to what is happening would be:

int i = 0;
int iAdd = i; // Copy of the current value of i, for ++
int iAssign = i; // Copy of the current value of i, for +=

i = i + 1; // i++ - Happens before += due to order of precedence
i = iAdd + iAssign;

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