Find the output of the following program
class Sample
{
public:
int *ptr;
Sample(int i)
{
ptr = new int(i);
}
~Sample()
{
delete ptr;
}
void PrintVal()
{
cout << "The value is " << *ptr;
}
};
void SomeFunc(Sample x)
{
cout << "Say i am in someFunc " << endl;
}
int main()
{
Sample s1= 10;
SomeFunc(s1);
s1.PrintVal();
}
Answer:
Say i am in someFunc
Null pointer assignment(Run-time error)
Explanation:
As the object is passed by value to SomeFunc the destructor of the object
is called when the control returns from the function. So when PrintVal is called
it meets up with ptr that has been freed.The solution is to pass the
Sample object by reference to SomeFunc:
void SomeFunc(Sample &x)
{
cout << "Say i am in someFunc " << endl;
}
because when we pass objects by refernece that object is not destroyed. while
returning from the function.
Which is the parameter that is added to
every non-static member function when it is called?
Answer:
‘this’
pointer
Find the output of the following program
class base
{
public:
int bval;
base(){ bval=0;}
};
class deri:public base
{
public:
int dval;
deri(){ dval=1;}
};
void SomeFunc(base *arr,int size)
{
for(int i=0; i
cout<bval;
cout<}
int main()
{
base BaseArr[5];
SomeFunc(BaseArr,5);
deri DeriArr[5];
SomeFunc(DeriArr,5);
}
Answer:
00000
01010
Explanation:
The function SomeFunc expects two arguments.The first one is a pointer to an
array of base class objects and the second one is the sizeof the array.The first
call of someFunc calls it with an array of bae objects, so it works correctly
and prints the bval of all the objects. When Somefunc is called the second time
the argument passed is the pointeer to an array of derived class objects and not
the array of base class objects. But that is what the function expects to be
sent. So the derived class pointer is promoted to base class pointer and the
address is sent to the function. SomeFunc() knows nothing about this and just
treats the pointer as an array of base class objects. So when arr++ is met, the
size of base class object is taken into consideration and is incremented by
sizeof(int) bytes for bval (the deri class objects have bval and dval as members
and so is of size >= sizeof(int)+sizeof(int) ).
Find the output of the following program
class base
{
public:
void baseFun(){
cout<<"from base"<
};
class deri:public base
{
public:
void baseFun(){
cout<< "from derived"<
};
void SomeFunc(base *baseObj)
{
baseObj->baseFun();
}
int main()
{
base baseObject;
SomeFunc(&baseObject);
deri deriObject;
SomeFunc(&deriObject);
}
Answer:
from base
from base
Explanation:
As we have
seen in the previous case, SomeFunc expects a pointer to a base class. Since a
pointer to a derived class object is passed, it treats the argument only as a
base class pointer and the corresponding base function is called.
Find the output of the following program
class base
{
public:
virtual void
baseFun(){ cout<<"from base"<
};
class deri:public base
{
public:
void baseFun(){
cout<< "from derived"<
};
void SomeFunc(base *baseObj)
{
baseObj->baseFun();
}
int main()
{
base baseObject;
SomeFunc(&baseObject);
deri deriObject;
SomeFunc(&deriObject);
}
Answer:
from base
from derived
Explanation:
Remember that
baseFunc is a virtual function. That means that it supports run-time
polymorphism. So the function corresponding to the derived class object is
called.
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