C++ Programming
SUBSCRIPTING MCQs
Total Questions : 9
The reason is that operator[] always takes exactly one parameter, but operator() can take any
number of parameters.
If you have a pointer to an object of some class type that overloads the subscript operator, you
have to dereference that pointer in order to free the memory.
What is the output of this program?
1.
#include
2.
using namespace std;
3.
class sample
4.
{
5.
private:
6.
int* i;
7.
int j;
8.
public:
9.
sample (int j);
10.
~sample ();
11.
int& operator [] (int n);
12.
};
13.
int& sample::operator [] (int n)
14.
{
15.
return i[n];
16.
}
17.
sample::sample (int j)
18.
{
19.
i = new int [j];
20.
j = j;
21.
}
22.
sample::~sample ()
23.
{
24.
delete [] i;
25.
}
26.
int main ()
27.
{
28.
sample m (5);
29.
m [0] = 25;
30.
m [1] = 20;
31.
m [2] = 15;
32.
m [3] = 10;
33.
m [4] = 5;
34.
for (int n = 0; n < 5; ++ n)
35.
cout
In this program, we are printing the array in the reverse order by using subscript operator.
Output:
$ g++ sub4.cpp
$ a.out
252015105
What is the output of this program?
1.
#include
2.
using namespace std;
3.
const int limit = 4;
4.
class safearray
5.
{
6.
private:
7.
int arr[limit];
8.
public:
9.
int& operator [](int n)
10.
{
11.
if (n == limit - 1)
12.
{
13.
int temp;
14.
for (int i = 0; i < limit; i++)
15.
{
16.
if (arr[n + 1] > arr[n])
17.
{
18.
temp = arr[n];
19.
arr[n] = arr[n + 1];
20.
arr[n + 1] = temp;
21.
}
22.
}
23.
}
24.
return arr[n];
25.
}
26.
};
27.
int main()
28.
{
29.
safearray sa1;
30.
for(int j = 0; j < limit; j++)
31.
sa1[j] = j*10;
32.
for(int j = 0; j < limit; j++)
33.
{
34.
int temp = sa1[j];
35.
cout
In this program, we are returning the array element by the multiple of 10.
Output:
$ g++ sub2.cpp
$ a.out
0102030
In this program, we overloading the operator[] by using subscript operator.
Output:
$ g++ sub3.cpp
$ a.out
001
What is the output of this program?
1.
#include
2.
using namespace std;
3.
class numbers
4.
{
5.
private:
6.
int m_nValues[10];
7.
public:
8.
int& operator[] (const int nValue);
9.
};
10.
int& numbers::operator[](const int nValue)
11.
{
12.
return m_nValues[nValue];
13.
}
14.
int main()
15.
{
16.
numbers N;
17.
N[5] = 4;
18.
cout
In this program, We are getting the values and returning it by overloading the subscript operator.
Output:
$ g++ sub1.cpp
$ a.out
4
None.
None.
The subscript operator overload takes only one argument,
but it can be of any type.
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