1) 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.
2)
Which is the parameter that is added to every
non-static member function when it is called?
Answer:
‘this’
pointer
3) 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<size;
i++,arr++)
cout<<arr->bval;
cout<<endl;
}
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) ).
4) class base
{
public:
void baseFun(){ cout<<"from base"<<endl;}
};
class deri:public base
{
public:
void
baseFun(){ cout<< "from derived"<<endl;}
};
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.
5) class base
{
public:
virtual
void baseFun(){ cout<<"from base"<<endl;}
};
class deri:public base
{
public:
void
baseFun(){ cout<< "from derived"<<endl;}
};
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.
6) void main()
{
int
a, *pa, &ra;
pa
= &a;
ra
= a;
cout
<<"a="<<a <<"*pa="<<*pa
<<"ra"<<ra ;
}
/*
Answer :
Compiler
Error: 'ra',reference must be initialized
Explanation :
Pointers
are different from references. One of the main differences is that the pointers can be
both initialized and assigned, whereas references can only be
initialized. So this code issues an error.
*/
7) const int size = 5;
void print(int *ptr)
{
cout<<ptr[0];
}
void print(int ptr[size])
{
cout<<ptr[0];
}
void main()
{
int
a[size] = {1,2,3,4,5};
int
*b = new int(size);
print(a);
print(b);
}
/*
Answer:
Compiler
Error : function 'void print(int *)' already has a body
Explanation:
Arrays
cannot be passed to functions, only pointers (for arrays, base addresses)
can be passed. So the arguments int *ptr
and int prt[size] have no difference
as function arguments. In other words,
both the functoins have the same signature and
so cannot be overloaded.
*/
8) class some{
public:
~some()
{
cout<<"some's
destructor"<<endl;
}
};
void main()
{
some
s;
s.~some();
}
/*
Answer:
some's
destructor
some's
destructor
Explanation:
Destructors
can be called explicitly. Here 's.~some()' explicitly calls the
destructor of 's'. When main() returns,
destructor of s is called again,
hence the result.
*/
9) #include <iostream.h>
class fig2d
{
int
dim1;
int
dim2;
public:
fig2d()
{ dim1=5; dim2=6;}
virtual
void operator<<(ostream & rhs);
};
void fig2d::operator<<(ostream
&rhs)
{
rhs
<<this->dim1<<" "<<this->dim2<<"
";
}
/*class fig3d : public fig2d
{
int
dim3;
public:
fig3d()
{ dim3=7;}
virtual
void operator<<(ostream &rhs);
};
void fig3d::operator<<(ostream
&rhs)
{
fig2d::operator
<<(rhs);
rhs<<this->dim3;
}
*/
void main()
{
fig2d
obj1;
// fig3d
obj2;
obj1
<< cout;
// obj2
<< cout;
}
/*
Answer :
5
6
Explanation:
In
this program, the << operator is overloaded with ostream as argument. This enables the 'cout' to be present at
the right-hand-side. Normally, 'cout' is implemented as global function, but
it doesn't mean that 'cout' is not possible to be overloaded as member function.
Overloading << as virtual member function becomes handy when the
class in which it is overloaded is inherited, and this
becomes available to be overrided. This is as opposed to global friend functions, where
friend's are not inherited.
*/
10) class opOverload{
public:
bool
operator==(opOverload temp);
};
bool opOverload::operator==(opOverload
temp){
if(*this == temp ){
cout<<"The
both are same objects\n";
return
true;
}
else{
cout<<"The
both are different\n";
return
false;
}
}
void main(){
opOverload
a1, a2;
a1=
=a2;
}
Answer :
Runtime
Error: Stack Overflow
Explanation :
Just
like normal functions, operator functions can be called recursively. This
program just illustrates that point, by calling the operator == function
recursively, leading to an infinite loop.
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