Find the output of the following program
void main()
{
int a, *pa, &ra;
pa = &a;
ra = a;
cout <<"a="<}
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.
Find the output of the following program
const int size = 5;
void print(int *ptr)
{
cout<}
void print(int ptr[size])
{
cout<}
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.
Find the output of the following program
class some{
public:
~some()
{
cout<<"some's destructor"<
}
};
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.
Find the output of the following program
#include
class fig2d
{
int dim1;
int dim2;
public:
fig2d() {
dim1=5; dim2=6;}
virtual void
operator<<(ostream & rhs);
};
void fig2d::operator<<(ostream &rhs)
{
rhs <dim1<<"
"<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<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.
Find the output of the following program
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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