Why Doesn\'t `std::queue::pop()` Return the Popped Element?

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

The lack of a return value in std::queue::pop() stems from the need for safety in the face of exceptions.

The original question highlights that inspecting an element from std::queue::front() still involves copying the element. However, the crucial difference lies in the potential for exceptions during the copying process.

Consider an implementation of std::queue::pop() that returns the popped element:

template<class T>
class queue {
    T pop() {
        auto x = elements[top_position];
        --top_position;  // alter queue state
        return x;        // calls T(const T&) which may throw
    }
};

If the copy constructor of T throws on return, the queue state has already been altered, but the element is not returned. This could result in data loss.

Moreover, such an implementation is inefficient if the popped value is not needed, as it creates an unnecessary copy.

Therefore, to ensure safety and efficiency, std::queue::pop() was designed as a void function, requiring clients to use std::queue::front() to inspect the value without creating a copy.

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