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Mar 23rd, 2006

homework question on class...multiple choice

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I have no clue what the answer is.

Which of the following dynamically allocates an object of type ourClass?

a) ourClass ourPtr;
ourPtr = new ourClass;

b) ourClass ourClassObj = new ourClass;

c) typedef ourClass* ourClassPtrType;
ourClassPtrType ourClassPtr;
ourClassPtr = new ourClassPtrType;

d) typedef ourClass* ourClassPtrType;
ourClassPtrType ourClassPtr;
ourClassPtr = new ourClassPtrType;

The class Derived is publicly derived from a class Base. The Base class has a public member function mem() that is not virtural. The member function mem() is redefined in class Derived.

Base *dPtr = new Derived;
dptr ->mem();

In the second line, which version of mem() is accessed?

I think the aswer is "b" but i'm not sure.

a) Base::mem()
b) Derived::mem()
c) This code causes an error
d) Both Base::mem() and Derived::mem()

any help will be appreciated.

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Mar 23rd, 2006

Re: homework question on class...multiple choice

>Which of the following dynamically allocates an object of type ourClass?
e, none of the above.

>In the second line, which version of mem() is accessed?
c, this code causes an error.

Narue

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Mar 23rd, 2006

Re: homework question on class...multiple choice

Compile this code.. then try changing bits of it... what happens when you modify it to some of the scenarios given in your homework assignment..?

C++ Syntax (Toggle Plain Text)
#include <iostream>
using namespace std;
 
class duck
{
public:
    virtual void speak() { cout << "quack"; }
};
 
class swan : public duck
{
public:
    void speak() { cout << "This is a swan"; }
};
 
class heron : public duck
{
public:
    void speak() { cout << "This is a heron"; }
};
 
int main()
{
    duck* myDuck = new swan;
    myDuck->speak();
    cin.get();
    return 0;
}
#include <iostream>
using namespace std;

class duck
{
public:
    virtual void speak() { cout << "quack"; }
};

class swan : public duck
{
public:
    void speak() { cout << "This is a swan"; }
};

class heron : public duck
{
public:
    void speak() { cout << "This is a heron"; }
};

int main()
{
    duck* myDuck = new swan;
    myDuck->speak();
    cin.get();
    return 0;
}

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Mar 23rd, 2006

Re: homework question on class...multiple choice

Quote originally posted by Narue ...

c, this code causes an error.

Why is that?

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Mar 24th, 2006

Re: homework question on class...multiple choice

Quote originally posted by sunnypalsingh ...

Why is that?

I think Narue is nitpicking a bit... nevertheless its the right answer according to the text in the post - The OP has probably made a typo (assuming he copied it from a book or sheet).. it took me a few reads to work out what the error was

(Compiler error would go something like "dptr: undeclared variable")

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Mar 24th, 2006

Re: homework question on class...multiple choice

oops~ sorry.

i made a typo on the first question....is that mean the answer is "C"

c) typedef ourClass* ourClassPtrType;
ourClassPtrType ourClassPtr;
ourClassPtr = new ourClass;

d) typedef ourClass* ourClassPtrType;
ourClassPtrType ourClassPtr;
ourClassPtr = new ourClassPtrType;

jack223

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Mar 24th, 2006

Re: homework question on class...multiple choice

>i made a typo on the first question
You made a typo on both questions.

>is that mean the answer is "C"
Yes, that's more like it.

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Mar 24th, 2006

Re: homework question on class...multiple choice

no, i don't think i made any mis typo on the second question...i double checked with my homework sheet...maybe the professor made a mistake.

Base *dPtr = new Derived;
dPtr -> mem();

In the second line, which version of mem() is accessed?

a) Base::mem()
b) Derived::mem()
c) This code causes an error.
d) Both Base::mem() and Derived::mem()

jack223

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Mar 24th, 2006

Re: homework question on class...multiple choice

>i don't think i made any mis typo on the second question
Then the worksheet had a typo. dptr is not the same as dPtr, and the example would fail to compile.

C++ Syntax (Toggle Plain Text)
Base *dPtr = new Derived;
dPtr -> mem();
Base *dPtr = new Derived;
dPtr -> mem();

A pointer has a real type and a virtual type. The real type of dPtr is Base (the type it was declared as), and the virtual type is Derived (the actual object type pointed to). Unless a function is declared as virtual, the real type will be used, so dPtr->mem() will call Base::mem since mem isn't virtual.

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Mar 24th, 2006

Re: homework question on class...multiple choice

if you look at my original question, i stated that "The Base class has a public member function mem() that is not virtural. The member function mem() is redefined in class Derived."

the base class mem() is not virtual so dPtr -> mem(); will call Base::mem??

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