How can I resolve type mismatch errors occurring in a `print_container` function?

Resolving Type Mismatch Errors in the print_container Function

To understand why the print_container function fails, we need to analyze the properties of the std::is_same template. This function compares two types and returns true if they are identical. In this case, we're using std::is_same to determine if the container type Cont is the same as a stack or a queue. However, the branches of the if-else statement depend on the exact type of the container, which leads to compilation errors since both branches must be compilable.

Solution 1: Partial Specialization

One solution is to use partial specialization to create separate implementations of element_accessor for stack and queue types. This allows us to handle the extraction of elements based on the specific container type. Here's an example implementation:

<code class="cpp">template <typename Cont>
struct element_accessor;

template <typename T>
struct element_accessor<std::stack<T>> {
   const T&amp; operator()(const std::stack<T>&amp; s) const { return s.top(); }
};

template <typename T>
struct element_accessor<std::queue<T>> {
   const T&amp; operator()(const std::queue<T>&amp; q) const { return q.front(); }
};

template<typename Cont>
void print_container(Cont&amp; cont){
   while(!cont.empty()){
      auto elem = element_accessor<Cont>{}(cont);
      std::cout << elem << '\n';
      cont.pop();
   }
}

In this solution, element_accessor provides a consistent way to access the top element for both stacks and front element for queues, resolving the type mismatch issue.

Solution 2: if constexpr (C 17)

If you're using C 17 or later, you can leverage the if constexpr syntax to implement branching based on template type parameters. Here's a modified version of the print_container function:

<code class="cpp">template<template<class> typename Cont, typename T>
void print_container(Cont<T>&amp; cont){
   while(!cont.empty()){
      if constexpr (std::is_same_v<Cont<T>, std::stack<T>>) 
         std::cout << cont.top() << '\n';
      else if constexpr (std::is_same_v<Cont<T>, std::queue<T>>) 
         std::cout << cont.front() << '\n';
      cont.pop();
   }
}</code>

In this version, the if constexpr branches are evaluated at compile-time based on the template type parameters. This eliminates the need for partial specialization and ensures type-safety.

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