需要对C ++中的Stacks和Templates进行一些澄清

So I have a problem I'm working on for my university class. Essentially the professor gives us a template in a header file and a main(), He asks us what would happen if we tried to "cout << s2.top();" and s2.pop(); after the multiple s2.push() and while loop (When the stack is empty). I already understand that after we cout the data in the stack, we pop() the last element until it is empty. Here is the code for stack.h:

所以我有一个问题,我正在为我的大学课程工作。基本上教授给我们一个头文件和一个main()的模板,他问我们如果我们试图“cout << s2.top();会发生什么?”和s2.pop();在多个s2.push()和while循环之后(当堆栈为空时)。我已经明白,在我们对堆栈中的数据进行cout之后,我们会弹出()最后一个元素,直到它为空。这是stack.h的代码:

#ifndef STACK_H
#define STACK_H
#include <vector>

using namespace std;

template <typename T>
class Stack
{
    vector<T> container;
public:
    Stack(): container() {}
    void push(T x) { container.push_back(x); }
    void pop() { if (container.size() > 0) container.pop_back(); }
    T top(){ if (container.size() > 0) return container.back(); }
    bool empty() { return container.empty(); }
};

#endif

Here is the main():

这是main():

#include <iostream>
#include <string>

#include "stack.h"

main()
{
Stack<int> s1;

s1.push(4);
s1.push(3);
s1.push(2);
s1.push(1);
while (!s1.empty()) {
    cout << s1.top() << endl;
    s1.pop();
}

Stack<string> s2;
s2.push("Yoda said ");
s2.push("something ");
s2.push("to write ");
while (!s2.empty()) {
    cout << s2.top();
    s2.pop();
}

s2.pop();
cout << s2.top();

cout << endl;

}

I know we will get a segmentation fault or something similar since were trying to access or pop() an empty stack. He wants us to ONLY change stack.h. I've already added:

我知道我们会在尝试访问或弹出()空堆栈时遇到分段错误或类似问题。他希望我们只改变stack.h。我已添加:

"void pop() { if (container.size() > 0) container.pop_back(); }"

“void pop(){if(container.size()> 0)container.pop_back();}”

And it will work for s2.pop().

它适用于s2.pop()。

My question is, when I try to add "if (container.size() > 0)" in the line:

我的问题是,当我尝试在行中添加“if(container.size()> 0)”时:

"T top(){ return container.back(); }"

“T top(){return container.back();}”

I still get a segmentation fault even though (at least I think) I'm checking the size before returning? How would i go about this one? Thanks in advance :)

即使(至少我认为)我在返回之前检查尺寸,我仍然会遇到分段错误?我怎么去这个?提前致谢 :)

2 个解决方案

#1

If control reaches } of a function returning non-void, instead of reaching return, the behaviour is undefined. Anything could happen.

如果控制到达返回非void的函数,而不是到达返回,则行为是未定义的。什么事情都可能发生。

You have to add else to the .top(), and use it to return some kind of default value (you probably want return {};), or throw an exception, or prevent the function from returning in some other way.

你必须在.top()中添加else,并使用它来返回某种默认值(你可能想要返回{};),或抛出异常,或阻止函数以其他方式返回。

#2

Your

s2.pop();
cout << s2.top();

outside the loop is a) calling pop() on an empty container (just silly). b) calling top() on an empty container and using the result; bug.

循环之外是a)在空容器上调用pop()(只是傻)。 b)在空容器上调用top()并使用结果;错误。

Also T top(){ if (container.size() > 0) return container.back(); } results in undefined behaviour when container is empty since then the function does not return a T (which it always must).

另外T top(){if(container.size()> 0)返回container.back();当容器为空时导致未定义的行为,因为该函数不返回T(它总是必须)。

#1