使用std :: is_same,为什么我的函数仍然不能用于2种类型 [英] using std::is_same, why my function still can't work for 2 types
问题描述
我正在尝试编写一个可以同时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 << '\n';
} else {
auto elem = cont.front();
std::cout << elem << '\n';
}
cont.pop();
std::cout << elem << '\n';
}
}
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;
}
但这在这里不起作用,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。但是我想知道为什么它不起作用的具体原因,以及如何处理它。
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.
PS:判断容器类型的实际含义是: :您可以通过传递队列容器而不是堆栈来简单地将DFS函数更改为BFS。因此,BFS和DFS可以共享大多数代码。
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.
PPS:我在C ++ 11环境中,但也欢迎使用较旧或更高标准的答案。
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 $的两个分支c $ c>语句必须是可编译的,这不是您的情况。基于局部专业化的许多可能解决方案之一,甚至可以在C ++ 98中使用。
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 << '\n';
cont.pop();
}
}
具有 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() << '\n';
else if constexpr (std::is_same_v<Cont<T>, std::queue<T>>)
std::cout << cont.front() << '\n';
cont.pop();
}
}
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