C++ Class Inheritance Segmentation Fault

Question

I'm doing a thing with C++ where I'm making an object of abstract class Parent and initializing it in a function to a child class. When I initialize the object when it's created, everything works wonders. When I initialize it in a function, I get a segmentation fault.

Here's a code outline of what I'm doing

_class.h

#ifndef CLASS_H_INCLUDED
#define CLASS_H_INCLUDED

class Parent{
  public:
    virtual void foo() = 0;
};

class Child1 : public Parent{
  public:
    virtual void foo(){
      ...
    }
};

class Child2 : public Parent{
  public:
    virtual void foo(){
      ...
    }
};

#endif

main.cpp

#include <iostream>

using namespace std;

#include "main.h"
#include "_class.h"

void setChild(Parent* c){
  c = new Child1();
}

int main() {
  Parent* c;// = new Child1();

  setChild(c);

  c->foo(); //Seg Fault
}

Obviously I could just initialize in main, but I'd rather initialize in the function. Is this possible, and if so, what am I doing wrong?

Parent* c passes a Parent by reference (specifically by a pointer), but the pointer itself is passed by value. c is a copy of the caller's argument and changes made to the copy are not duplicated at the caller. Use a reference : void setChild(Parent* & c). — user4581301
– user4581301, Commented Mar 5, 2020 at 16:29
Pass a Parent** p to setChild and assign the result of new Child1(); to *p. — JohnFilleau
– JohnFilleau, Commented Mar 5, 2020 at 16:30

1201ProgramAlarm · Accepted Answer · 2020-03-05 16:30:31Z

2

You're setting the local c in setChild to a new object, but the one in main is not changed. There are two ways to change the value in the caller.

One is to pass the parameter by reference:

void setChild(Parent*& c)

the other is to return it, rather than use a parameter

Parent *setChild() {
    return new Child1();
}

answered Mar 5, 2020 at 16:30

1201ProgramAlarm

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1 Comment

ObscurityGamer Over a year ago

Sometimes I hate C++. Defeated by an &. I'll accept your answer as soon as it allows me. Thank you.

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