Why does std::is_same fail when trying to process different container types in a function?-C++-php.cn

Why does std::is_same fail when trying to process different container types in a function?

Mary-Kate Olsen

Release： 2024-11-01 14:09:02

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Why does std::is_same fail when trying to process different container types in a function?

Using std::is_same: Why the Function Fails to Process Different Container Types

In an attempt to create a versatile function capable of printing both stacks and queues, a developer encounters a compilation error. Despite utilizing std::is_same to discern the container's type, the function remains non-functional.

The error originates from the fact that both branches of the if-else statement must compile successfully, which is violated in this instance. To address this issue, a modification is implemented using partial specialization and element_accessor template struct:

<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();
   }
}

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Alternatively, for C 17 and higher, the error can be circumvented using if constexpr rather than partial specialization:

<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>

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