Why Doesn\'t `std::queue::pop()` Return a Value?

Why std::queue::pop Doesn't Return a Value

Despite returning seemingly redundant information, std::queue::pop() deliberately omits a return value to ensure safety in the presence of exceptions.

The suggested solution of returning the popped element by reference (as in myqueue.front()) still requires a copy to be made at the point of usage. This copy operation, commonly known as a move or copy constructor, may throw an exception.

Consider the following scenario:

auto x = myqueue.pop(); // calls the copy constructor of T

If the copy constructor of T fails, the queue's state is already altered (e.g., the element is removed from the queue) but the return value is not produced. This leaves the queue in an inconsistent state and potentially loses the popped element.

Furthermore, returning a reference also introduces an efficiency issue when the popped value is not needed. In contrast, the current design of std::queue separates the operations of removing an element (pop()) and inspecting its value (front()), providing both safety and efficiency.

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