Why does cout << not print inputs from left to right?
c++, cout, order-of-execution, sequence-points
Solution
There's no sequence point with the `<<` operator so the compiler is free to evaluate either `dequeue` function first. What is guaranteed is that the result of the second `dequeue` call (in the order in which it appears in the expression and not necessarily the order in which it is evaluated) is `<<`'ed to the result of `<<`'ing the first (if you get what I'm saying).
So the compiler is free to translate your code into some thing like any of these (pseudo intermediate c++). This isn't intended to be an exhaustive list.
auto tmp2 = myQueue.dequeue();
auto tmp1 = myQueue.dequeue();
std::ostream& tmp3 = cout << tmp1;
tmp3 << tmp2;
or
auto tmp1 = myQueue.dequeue();
auto tmp2 = myQueue.dequeue();
std::ostream& tmp3 = cout << tmp1;
tmp3 << tmp2;
or
auto tmp1 = myQueue.dequeue();
std::ostream& tmp3 = cout << tmp1;
auto tmp2 = myQueue.dequeue();
tmp3 << tmp2;
Here's what the temporaries correspond to in the original expression.
cout << myQueue.dequeue() << myQueue.dequeue();
| | | | |
| |____ tmp1 _____| |_____ tmp2 ____|
| |
|________ tmp3 _________|
Problem
the following code: ``` myQueue.enqueue('a'); myQueue.enqueue('b'); cout << myQueue.dequeue() << myQueue.dequeue(); ``` prints "ba" to the console while: ``` myQueue.enqueue('a'); myQueue.enqueue('b'); cout << myQueue.dequeue(); cout << myQueue.dequeue(); ``` prints "ab" why is this? It seems as though cout is calling the outermost (closest to the ;) function first and working its way in, is that the way it behaves?