浅拷贝不崩溃在深入研究C++的同时，我正在进行浅/深的复制。根据我跟踪的MOOC，以下代码应该崩溃 // .hpp file #ifndef _COPY_HPP_ #define __COPY_HPP_ class Shallow { private: int* data; public: void setData (int d){ *data = d;} int getData(){ return *data;} // constructor Shallow (int d); // copy constructor Shallow(const Shallow &source); //destructor ~Shallow(); }; #endif // .cpp file #include "copy.hpp" #include<stdio.h> #include<iostream> using namespace std; Shallow::Shallow(int d){ data = new int; *data = d; } void displayShallow(Shallow s){ cout << "displayShallow: " << s.getData() << "\n"; } Shallow::Shallow(const Shallow &source) : data{source.data} { cout << "copy constructor call" << "\n"; } Shallow::~Shallow(){ delete data; cout << "freeing data" << "\n"; } int main(){ Shallow original{100}; cout << original.getData() << "\n"; // destructor called after this call displayShallow(original); cout << original.getData() << "\n"; original.setData(29); cout << original.getData() << "\n"; return 0; }_C++

浅拷贝不崩溃在深入研究C++的同时，我正在进行浅/深的复制。根据我跟踪的MOOC，以下代码应该崩溃 // .hpp file #ifndef _COPY_HPP_ #define __COPY_HPP_ class Shallow { private: int* data; public: void setData (int d){ *data = d;} int getData(){ return *data;} // constructor Shallow (int d); // copy constructor Shallow(const Shallow &source); //destructor ~Shallow(); }; #endif // .cpp file #include "copy.hpp" #include<stdio.h> #include<iostream> using namespace std; Shallow::Shallow(int d){ data = new int; *data = d; } void displayShallow(Shallow s){ cout << "displayShallow: " << s.getData() << "\n"; } Shallow::Shallow(const Shallow &source) : data{source.data} { cout << "copy constructor call" << "\n"; } Shallow::~Shallow(){ delete data; cout << "freeing data" << "\n"; } int main(){ Shallow original{100}; cout << original.getData() << "\n"; // destructor called after this call displayShallow(original); cout << original.getData() << "\n"; original.setData(29); cout << original.getData() << "\n"; return 0; }

c++

浅拷贝不崩溃在深入研究C++的同时，我正在进行浅/深的复制。根据我跟踪的MOOC，以下代码应该崩溃 // .hpp file #ifndef _COPY_HPP_ #define __COPY_HPP_ class Shallow { private: int* data; public: void setData (int d){ *data = d;} int getData(){ return *data;} // constructor Shallow (int d); // copy constructor Shallow(const Shallow &source); //destructor ~Shallow(); }; #endif // .cpp file #include "copy.hpp" #include<stdio.h> #include<iostream> using namespace std; Shallow::Shallow(int d){ data = new int; *data = d; } void displayShallow(Shallow s){ cout << "displayShallow: " << s.getData() << "\n"; } Shallow::Shallow(const Shallow &source) : data{source.data} { cout << "copy constructor call" << "\n"; } Shallow::~Shallow(){ delete data; cout << "freeing data" << "\n"; } int main(){ Shallow original{100}; cout << original.getData() << "\n"; // destructor called after this call displayShallow(original); cout << original.getData() << "\n"; original.setData(29); cout << original.getData() << "\n"; return 0; },c++,C++,似乎数据指向的存储被设置为0，而不是崩溃。我是否正确理解？< /p>“未定义的行为”是指“它可能崩溃，它可能不会崩溃，它可能释放愤怒”，根据MOOC——什么是“MOOC”？这是另一个在线C++课程给出错误信息。这样的保证有什么用？通常的误解是，在C++中调用UB将强制导致崩溃。不幸的是，情况并非如此。这很好，但它会花费成本，C++有范例——不为你不使用的东西付费。 100 copy constructor call displayShallow: 100 freeing data 0 29 fr

似乎

数据

指向的存储被设置为

，而不是崩溃。我是否正确理解？< /p>“未定义的行为”是指“它可能崩溃，它可能不会崩溃，它可能释放愤怒”，根据MOOC——什么是“MOOC”？这是另一个在线C++课程给出错误信息。这样的保证有什么用？通常的误解是，在C++中调用UB将强制导致崩溃。不幸的是，情况并非如此。这很好，但它会花费成本，C++有范例——不为你不使用的东西付费。

100
copy constructor call
displayShallow: 100
freeing data
0
29
freeing data