Order of execution in operator <<

Question

I have difficulties in understanding the sequence of calls in the code below. I was expecting to see the output below

    A1B2

While I can

Answer 1

The shift operators are left-associative; a << b << c is read as (a << b) << c, meaning that if a is of a type with member user-defined operator<< (and returns that type) then the expression reads as a.operator<<(b).operator<<(c). If instead a free operator<< is used, then this reads as operator<<(operator<<(a, b), c).

So the evaluation of a << b is sequenced before the evaluation of (a << b) << c, but there is no sequencing dependency between the evaluation of b and c:

a << b << c[1]
|         |
a << b[2] |
|    |    c[5]
a[3] b[4]

If we number the side-effects as above, then the side-effects can be sequenced as any of:

54321
53421
45321
43521
43251
35421
34521
34251