C ++自定义堆栈打印问题

Question

I am creating a custom class Stack to store a couple string variables. When I attempt to print the stack, it says that the stack is always empty, which is not correct. I am using vectors to represent the custom stack, so the way I go about my print method should work, but for some reason it does not. What is my error? Is it in my isEmpty method?

void stack::printStack() {

    std::vector<std::string> v;

    if(stack::isEmpty()) {

        std::cout << "Stack is empty! " << std::endl;
    }
    else {
        for(int i = 0; i != v.size(); i++) {
            std::cout << v[i] << std::endl;
        }
    }
}

Answer 1

You are returning a copy of the underlying vector with

std::vector<std::string> stack::getVector();

This results in all calls such as stack::getVector().push_back(n); making no modifications to the stack::v , but modifying the returned temporary instead.

I don't see, why you shouldn't be using v directly in the member functions:

v.push_back(n);

Or, if you don't want to do that (for some reason), make getVector return both reference-to-const and reference-to-non-const:

std::vector<std::string> const& stack::getVector() const { return v; };
std::vector<std::string> &stack::getVector()             { return v; };

Note, that you're breaking encapsulation with std::vector<std::string> & returning overload.

Answer 2

You are creating a local variable here:

void stack::printStack(){

      std::vector<std::string> v;

That v is not the same as the v member variable. That local v is empty, thus your loop never prints.

Also, use descriptive variable names. Using a single letter variable name such as v is not a good idea.

Also, with respect to return a vector by reference or copy, see this question and answer: Returning vector copies.

So do you want to return a copy of the vector, or do you want to return the actual vector? If it is the latter, return a reference, if it's the former, then return a copy (as your current code is doing). Note that there are implications in returning a copy as opposed to returning a reference (as the link shows).