当堆栈为空时，在(模板化)堆栈弹出方法中该怎么办? [英] What to do in a (templated) stack pop method when the stack is empty?

问题描述

我已经编写了一个模板化的容器类，该类采用模板参数的类型和模板.

I have written a templatized container class that takes type and template of template parameter.

template<class type, template<typename...> class Seq>
class stack1
{
private:
    int count;
    int size;
    Seq<type> st;
    //Seq<string> str;

public:
    stack1(size_t size):size(100), count(-1){ }
    void push(type elem);
    type pop();

};


template<class type, template<typename...> class Seq>
type stack1<type, Seq>::pop()
{
    if (count < 0)
    {
        cout << "stack1 is empty," << endl;
        /*How to handle this condition.*/

    }
    else
    {
        type elem; 
        elem = st.back();
        st.pop_back();
        count--;
        return elem;
    }
}

我的问题是，在pop函数中，当容器对象为空时，我应该如何处理错误情况.在这种情况下，我想返回一些默认值，例如如果容器为int，则为0/-1；如果为字符串，则为"/null；如果为float，则为0.0.

my question is , in the pop function how should I handle the error scenario when the container object is empty. I want to return some default value in that case, e.g. 0/-1 if the container is int or ""/null if it is string or 0.0 in case it is float... something like that.

推荐答案

@RSahu的建议很好.

@RSahu's suggestion is a fine thing to have.

另一种选择是将 pop()函数的签名从以下位置更改:

An alternative could be changing the signature of the pop() function from:

type pop();

到

std::optional<type> pop();

，如果堆栈为空，则返回 std :: nullopt ，或者在通常情况下返回包装值:

and returning std::nullopt if the stack is empty, or a wrapped value in the usual case:

if (count < 0) {
    return std::nullopt;
}

请注意，在C ++ 17语言标准中引入了 std :: optional ；在C ++ 14中，您可以将其作为 std :: experimental :: optional ，或者对于C ++ 11和更早版本，可以使用 boost :: optional .

Note that std::optional is introduced in the C++17 language standard; in C++14 you have it as std::experimental::optional, or you can use boost::optional for C++11 and earlier.

PS:当元素计数实际上为0时，将count设为-1是一个糟糕的主意-非常令人困惑！

PS: It's a bad idea to have count be -1 when the element count is actually 0 - very confusing!

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