C++ - 返回对象的函数

Question

// Assume class definition for Cat is here.

Cat makeCat() {
    Cat lady = new Cat("fluffy");
    return lady;
}

int main (...) {
    Cat molly = makeCat();
    molly->eatFood(); 
    return 0;
}

是否会出现"免费使用"错误molly->eatFood()？

Answer 1

更正了您的程序并创建了一个示例实现class Cat:

#include <iostream>
#include <string>

class Cat {
public:
        Cat(const std::string& name_ = "Kitty")
        : name(name_)
        {
                std::cout << "Cat " << name << " created." << std::endl;
        }
        ~Cat(){
                std::cout << "Cat " << name << " destroyed." << std::endl;
        }
        void eatFood(){
                std::cout << "Food eaten by cat named " << name << "." << std::endl;
        }
private:
        std::string name;
};

Cat* makeCat1() {
        return new Cat("Cat1");
}

Cat makeCat2() {
        return Cat("Cat2");
}

int main (){
        Cat kit = makeCat2();
        kit.eatFood();

        Cat *molly = makeCat1();
        molly->eatFood();
        delete molly;

        return 0;
}

它会产生输出:

Cat Cat2 created.
Food eaten by cat named Cat2.
Cat Cat1 created.
Food eaten by cat named Cat1.
Cat Cat1 destroyed.
Cat Cat2 destroyed.

我建议你在继续之前学习一本关于C++封面的基础书.

Answer 2

new Cat("fluffy")创建一个指针.您需要指定Cat*为返回类型.由于对象是在堆中创建的,因此在返回函数后它仍然可用.

Answer 3

除了在程序结束时发生内存泄漏外，对于内存的无效使用没有错误。如果在堆上创建了某些东西（例如使用new），则需要调用delete它以释放它。

您也有很多语法错误，下面将进行更正。

Cat* makeCat() 
{
  Cat *lady = new Cat("fluffy");
  return lady;
}

int main (int argc, char** argv) 
{

  Cat* molly = makeCat();
  molly->eatFood(); 

  delete molly;//This was added
  return 0;
}