Question

Trace this code:

1)

#include<iostream>

using namespace std;

  

class Test {

    int value;

public:

    Test(int v);

};

  

Test::Test(int v) {

    value = v;

}

  

int main() {

    Test t[100];

    return 0;

}

===================================================================

2)

#include <iostream>

using namespace std;

int main()

{

                int i, j;

                for (i = 1; i <= 3; i++)

                {

                                //print * equal to row number

                                for (j = 1; j <= i; j++)

                                {

                                                cout << "* ";

                                }

                                cout << "\n";

                }

                system("pause");

                return 0;

Answer 1

1)

Note:- In main() the first statement should be Test t(100); not Test t[100];

If it is Test t[100] then the program will return compile time error.

#include<iostream>
using namespace std;
class Test {
int value;
public:
Test(int v);
};
Test::Test(int v) { // definition of Test() which is a Test class method
value = v; // cout<<value; => it prints value 100 that is passed from main()
}
int main() {
Test t(100);
return 0;
}

Here, In this code a class named Test is declared and an integer value and a public function Test() with parameter integer value are declared. In main() we pass a value of 100 by creating object for Test class. So the value attribute in Test class is assigned a value of 100.

2) #include <iostream>
using namespace std;
int main()
{
   int i, j;
for (i = 1; i <= 3; i++){
       for (j = 1; j <= i; j++){
           cout << "* ";
       }
cout << "\n";
   }
   system("pause"); // It tells the operating system to pause program and waits until it is terminated manually.
   return 0;
}

Tracing:- for i=1 => j=1 => only one * is printed and enters next line.

for i=2 => j=1,2 => two * ' s separated by space are printed and enters next line.

for i=3 => j=1,2,3 => three * ' s separated by space are printed and enters next line.

So the output will be

*

* *

* * *