Question

Show what is output by the following C++ code if infile contains: 3 4 5 3 -1 12 8 a =3; infile >> x; while (x > 0) { a = a + x; cout << a << endl; infile >> x; }
C++ code

Answer 1

a=3;
infile>>x;
while(x>0)
{
a=a+x;
cout<<a<<endl;
infile>>x;
}

The optput of above code will be

6
10
15
18

It can be explained as: In first iteration the value of x is 3 so a=6(3+3)
Then in second iteration x=4 and a=6
so, a=6+4=10

Third iteration x=5 and a=10
so, a=10+5=15

Fourth iteration: x=3 nd a=15
so, a=15+3=18

In fifth iteration the value of x is -1 hich is less than 0. So, the condition of while loop(x>0) will not be satisfied and the counter will come out of the loop.