如何从队列中获取前5个元素C ++

Question

如何在不使用for循环的情况下从队列中获取前五个元素？

Answer 1

No explicit loops at all:

#include <iostream>
#include <queue>
#include <vector>
#include <algorithm>

template <typename T>
class QueuePopper
{
public:
    QueuePopper(std::queue<T> &q) : q(q) {}
    T operator() (const T &) { T t = q.front(); q.pop(); return t; }
private:
    std::queue<T> &q;
};

int main()
{
    typedef std::string T;

    std::queue<T> q;

    ...

    std::vector<T> v(5);
    std::transform(v.begin(), v.end(), v.begin(), QueuePopper<T>(q));
}

Answer 2

One more way:

template<int n> void GetElements(queue<string>& qu)
{
    if(!qu.empty())
    {
        cout << qu.front() << '\n';
        qu.pop();
    }
    GetElements<n-1>(qu);
}

template<> void GetElements<0>(queue<string>& qu) {}

// and call:
GetElements<5>(qu);

Or:

switch(qu.size())
{
default: cout << qu.front() << '\n'; qu.pop();
case 4: cout << qu.front() << '\n'; qu.pop();
case 3: cout << qu.front() << '\n'; qu.pop();
case 2: cout << qu.front() << '\n'; qu.pop();
case 1: cout << qu.front() << '\n'; qu.pop();
case 0: ;
}

Answer 3

Why would you want to do that? Maybe you have to hold 5 elements in hand at a time fr reasons that are not clear here.

Anyway - if you means STL <queue> :

std::queue<Element> container;

Element e1, e2, e3, e3, e5;

e1 = container.front();
container.pop();
e2 = container.front();
container.pop();
e3 = container.front();
container.pop();
e4 = container.front();
container.pop();
e5 = container.front();
container.pop();

Behaviour is undefined if < 5 elements on the queue .

You can replace this with the below if you want to avoid the Element copy, however in this case be aware that the reference becomes invalid after the referenced element is popped.

std::queue<Element> container;

Element& e1 = container.front();    // or const Element&
container.pop();
Element& e2 = container.front();
container.pop();
Element& e3 = container.front();
container.pop();
Element& e4 = container.front();
container.pop();
Element& e5 = container.front();
container.pop();

Answer 4

#include <iostream>
#include <queue>

int main(){
    std::queue<std::string> q;
    for (int i = 0; i<5; ++i){
        std::string s;
        std::cin >> s;
        q.push(s);
    }
    int itemsToFetch = 5;
    do{
        std::string curString = q.front();
        q.pop();
        std::cout << curString << std::endl;
        --itemsToFetch;
    } while (itemsToFetch > 0 && !q.empty());
    return 0;
}