Double assignment of the same variable in one expression in C++11

Question

The C++11 standard (5.17, expr.ass) states that

In all cases, the assignment is sequenced after the value computation of the right and left operands, a

Answer 1

Let's rewrite your code as

E1 = (E2 = E3)

where E1 is the expression a, E2 is the expression a += 1 and E3 is the expression 10. Here we ussed, that the assignment operator groups right-to-left (§5.17/1 in C++11 Standard).

§5.17/1 moreover states:

In all cases, the assignment is sequenced after the value computation of the right and left operands, and before the value computation of the assignment expression.

Applying this to our expression means that we first must evaluate the subexpressions E1 and E2 = E3. Note that there is no "sequenced-before" relationship between these two evaluations, but that causes no problems.

The evaluation of the id-expression E1 is trivial (the result is a itself). The evaluation of the assignment-expression E2 = E3 proceeds as follows:

First both subexpressions have to be evaluated. The evaluation of the literal E3is again trivial (gives a prvalue of value 10).

The evaluation of the (compound) assignment-expression E2 is done in the following steps:

1) The behavior of a += 1is equivalent to a = a + 1 but a is only evaluated once (§5.17/7). After evaluating the subexpressions a and 1 (in an arbitrary order), an lvalue-to-rvalue conversion is applied to a in order to read the value stored in a.

2) The values of a (which is 0) and of 1 are added (a + 1) and the result of this addition is a prvalue of value 1.

3) Before we can compute the result of the assignment a = a + 1 the value of the object the left operand refers to is replaced by the value of the right operand (§5.17/2). The result of E2 is then an lvalue refereing to the new value 1. Note that the side effect (updating the value of the left operand) is sequenced before the value computation of the assignment expression. This is §5.17/1 cited above.

Now that we have evaluated the subexpressions E2and E3, the value of the expression E2refers to is replaced by the value of E3, which is 10. Hence the result of E2 = E3 is an lvalue of value 10.

Finally, the value expression E1 refers to is replaced by the value of the expression E2 = E3, which we computed to be 10. Thus, the variable aends up to contain the value 10.

Since all these steps are well-defined, the whole expression yields a well-defined value.