使用 std::is_same，为什么我的函数仍然不能用于 2 种类型
我正在尝试编写一个可以打印堆栈和队列的函数，我的代码如下
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;
}
但是这里不起作用，错误信息是:
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.
我知道这是有问题的，并且知道 C++ 是静态类型的并且没有太多的运行时支持.但我想知道这不起作用的具体原因，以及如何处理.
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.:判断容器类型的实际含义是:你可以简单地通过传递队列容器而不是堆栈来将DFS函数更改为BFS.因此，BFS 和 DFS 可以共享大部分代码.
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 环境中，但也欢迎对旧版或更新版标准的回答.
P.P.S: I am in C++ 11 environment, but answers for older or later standard are also welcomed.
推荐答案
if-else 语句的两个分支都必须是可编译的，而在您的情况下则不然.基于部分专业化的许多可能的解决方案之一，即使在 C++98 中也应该工作:
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();
}
}
<小时>
带有 if constexpr 的 C++17 解决方案:
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();
}
}
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