Why Does the Order of `cout

Operator Evaluation Order in C Operator Expressions

In the given code snippets:

myQueue.enqueue('a');
myQueue.enqueue('b');
cout << myQueue.dequeue() << myQueue.dequeue();

and

myQueue.enqueue('a');
myQueue.enqueue('b');
cout << myQueue.dequeue();
cout << myQueue.dequeue();

the ordering of the print statements might seem unexpected. However, understanding the evaluation order of the << operator clarifies this behavior.

The << operator has no defined evaluation order for its arguments. This means the compiler is free to evaluate them in any sequence. However, it guarantees that the second argument (the one to the right of the <<) is evaluated after the first.

In the first code snippet, the order of evaluation doesn't matter because both dequeue() calls retrieve different elements from the queue. However, in the second snippet, the order is crucial.

The compiler can evaluate the code in various orders, including:

auto tmp1 = myQueue.dequeue();
auto tmp2 = myQueue.dequeue();
cout << tmp1 << tmp2;

and

auto tmp1 = myQueue.dequeue();
cout << tmp1 << myQueue.dequeue();

The first ordering will output "ab", while the second outputs "ba". The specific output depends on the order of evaluation chosen by the compiler.

This behavior emphasizes the importance of understanding the evaluation order of operators when writing C code, especially when dealing with operators that have undefined or unsequenced evaluation order.

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