問題描述

我正在嘗試編寫一個可以打印堆棧和隊列的函數，我的代碼如下

I am trying to write a function that can print both stack and queue, my code is as following

template<typename Cont>
void print_container(Cont& cont){
    while(!cont.empty()){
        if(std::is_same<Cont, stack<int>>::value){
            auto elem = cont.top();
            std::cout << elem << '
';
        } else {
            auto elem = cont.front();
            std::cout << elem << '
';
        }
        cont.pop();
        std::cout << elem << '
';
    }
}

int main(int argc, char *argv[])
{
    stack<int> stk;
    stk.push(1);
    stk.push(2);
    stk.push(3);
    queue<int> q;
    q.push(1);
    q.push(2);
    q.push(3);

    std::cout << "print stack" << endl;
    print_container(stk);
    std::cout << "print queue" << endl;
    print_container(q);

    return 0;
}

但是這里不起作用，錯誤信息是:

demo_typeof.cpp:35:30: error: no member named 'front' in 'std::__1::stack<int, std::__1::deque<int, std::__1::allocator<int> > >'
            auto elem = cont.front();
                        ~~~~ ^
demo_typeof.cpp:52:5: note: in instantiation of function template specialization 'print_container<std::__1::stack<int, std::__1::deque<int, std::__1::allocator<int> > > >' requested here
    print_container(stk);
    ^
demo_typeof.cpp:32:30: error: no member named 'top' in 'std::__1::queue<int, std::__1::deque<int, std::__1::allocator<int> > >'
            auto elem = cont.top();
                        ~~~~ ^
demo_typeof.cpp:54:5: note: in instantiation of function template specialization 'print_container<std::__1::queue<int, std::__1::deque<int, std::__1::allocator<int> > > >' requested here
    print_container(q);
    ^
2 errors generated.

我知道這是有問題的，并且知道 C++ 是靜態類型的并且沒有太多的運行時支持.但我想知道這不起作用的具體原因，以及如何處理.

I know it's problematic, and know C++ is statically typed and without too much Runtime support. but I am wondering the specific reason why this doesn't work, and how to deal with it.

P.S.:判斷容器類型的實際含義是:你可以簡單地通過傳遞隊列容器而不是堆棧來將DFS函數更改為BFS.因此，BFS 和 DFS 可以共享大部分代碼.

P.S.: The actual meaning of judge the typing of a container is that: You can simply change a DFS function into BFS by passing a queue container instead of a stack. So, BFS and DFS can share most of the code.

P.P.S:我在 C++ 11 環境中，但也歡迎對舊版或更新版標準的回答.

P.P.S: I am in C++ 11 environment, but answers for older or later standard are also welcomed.

推薦答案

if-else 語句的兩個分支都必須是可編譯的，而在您的情況下則不然.基于部分專業化的許多可能的解決方案之一，即使在 C++98 中也應該工作:

Both branches of if-else statement must be compilable, which are not in your case. One of many possible solutions that is based on partial specialization and should work even in C++98:

template <typename Cont>
struct element_accessor;

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

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

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

<小時>

帶有 if constexpr 的 C++17 解決方案:

A C++17 solution with if constexpr:

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

這篇關于使用 std::is_same，為什么我的函數仍然不能用于 2 種類型的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！