如何将名称传递给基类的构造函数

Question

I have two classes and I want every object of subclass to have a name, which is passed as an argument to the constructor of superclass . How do I do this?

class base
{
   public:
     base(const char * name) {name = name;}
     const char * getName() {return name;};
   private:
     const char * name;
};
class derived : public base
{
   public:
     derived(const char * name) : base(name) {}
};

However,

int main(int argc, char **argv)
{
    derived d("test");
    std::cout << d.getName();
}

produces gibberish as the the console output.

Answer 1

You need to choose better name for parameter name :

base(const char * p_name) {name = p_name;}

in you code you are just assigning name to itself.

Other possible solutions:

base(const char * name) : name(name) {} // preferred one
base(const char * name) {base::name = name;}
base(const char * name) {this->name = name;}

Answer 2

The problem occurs because you're assigning name to itself. So the member variable const char * name won't have the correct value. Since this is C++ though you should really use std::string :

#include <string>
class base
{
   public:
     base(std::string name) {m_name = name;}
     std::string getName() {return m_name;};
   private:
     std::string m_name;
};
class derived : public base
{
   public:
     derived(std::string name) : base(name) {}
};

Answer 3

Your problem is the way you use name in the body of the constructor, where it is local parameter. To fix the code, either use initialization lists or refer to member name through this .

Example 1 (member list, preferred!):

base(const char * name) : name(name) {}

Example 2 (use this , not preferred):

base(const char * name) : { this->name = name; }

Answer 4

• name is used twice in the constructor, you could change the member to name_ for example:

   class base { public: base(const char * name) {name_ = name;} const char * getName() {return name_;}; private: const char * name_; }; 

  or use a initialization list:

    base(const char * name) : name(name) {} 

  if you want to keep the same name for the parameter as well as for the member variable.

• In C++, you can use std::string instead of a char* for the name.

• In class base , no memory is allocated for name (in this case it would be better to copy the data with strcpy() instead of keeping a pointer to a constant parameter).

Answer 5

You have a problem with ambiguous use of the name symbol.

The obvious answer is:

Make the symbols of parameter and member variable unambiguous.

---

Your code can be fixed like this:

#include <iostream>

class base
{
   public:
     base(const char * name) {name_ = name;}
     const char * getName() {return name_;};
   private:
     const char * name_; // <<< Just make the member variable unambiguous.
};
class derived : public base
{
   public:
     derived(const char * name) : base(name) {}
};

int main(int argc, char **argv)
{
    derived d("test");
    std::cout << d.getName() << std::endl;
}

See the working demo please.

Answer 6

base(const char * name) {name = name;}

is not doing what you are expecting. In the function, the argument name shadows the member variable name . Hence, the member variable never gets initialized.

You can use:

base(const char * name) : name(name) {}

or better

base(const char * nameIn) : name(nameIn) {}

Suggestion for further improvement:

Use a std::string for the member variable. Then your class doesn't have to deal with problems that might arise if the pointer becomes a dangling pointer after the constructor is called.

class base
{
   public:
     base(const char * nameIn) : name(nameIn) {}
     std::string const& getName() {return name;} const;
   private:
     std::string name;
};