MCQs
Total Questions : 5
Answer: Option A. -> It combines the first operand and the second operand
The binary operator .* combines its first operand, which must be an object of class type, with
its second operand, which must be a pointer-to-member type.
Answer: Option A. -> Interface
None.
Answer: Option A. -> Executed
In this program, We passes the value to the class and printing it.
Output:
$ g++ ptm4.cpp
$ a.out
Executed
Question 4.
1.
#include
2.
using namespace std;
3.
class bowl
4.
{
5.
public:
6.
int apples;
7.
int oranges;
8.
};
9.
int count_fruit(bowl * begin, bowl * end, int bowl :: *fruit)
10.
{
11.
int count = 0;
12.
for (bowl * iterator = begin; iterator != end; ++ iterator)
13.
count += iterator ->* fruit;
14.
return count;
15.
}
16.
int main()
17.
{
18.
bowl bowls[2] = {{ 1, 2 },{ 3, 5 }};
19.
cout
What is the output of this program?
1.
#include
2.
using namespace std;
3.
class bowl
4.
{
5.
public:
6.
int apples;
7.
int oranges;
8.
};
9.
int count_fruit(bowl * begin, bowl * end, int bowl :: *fruit)
10.
{
11.
int count = 0;
12.
for (bowl * iterator = begin; iterator != end; ++ iterator)
13.
count += iterator ->* fruit;
14.
return count;
15.
}
16.
int main()
17.
{
18.
bowl bowls[2] = {{ 1, 2 },{ 3, 5 }};
19.
cout
Answer: Option A. -> Executed
Answer:a
Explanation:In this program, We are passing the value to the class and adding the values and printing it in the main.
Output:
$ g++ ptm3.cpp
$ a.out
I have 4 apples
I have 7 oranges
Answer: Option C. -> Both a & b
In this program, We are printing the value by direct access and another one by using pointer
to member.
Output:
$ g++ ptm2.cpp
$ a.out
1
2