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C++ Looping | Find output programs | Set 5

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robaa
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2022-11-21
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This section contains the C++ find output programs with their explanations on C++ Looping (set 5).

Program 1:

#include <iostream>
using namespace std;

int main()
{
    int num = 15673;
    int R1 = 0, R2 = 0;

    do {
        R1 = num % 10;
        R2 = R2 * 10 + R1;

        num = num / 10;
    } while (num > 0);

    cout << R2 << " ";

    return 0;
}

Program 2:

#include <iostream>
using namespace std;

int main()
{
    int I = 1;
    int D = 0;
    int R = 0;

    do {
        R = I++ * D++;

        cout << R << " ";
    } while (I <= 5);

    return 0;
}

Program 3:

#include <iostream>
using namespace std;

int main()
{
    int I = 1;
    int D = 1;
    int R = 0;

    do {
        R = I++ * D++;
        if (I == 3)
            continue;
        cout << R << " ";
    } while (I <= 5);

    return 0;
}
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All Answers

robaa
2022-11-21
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Answer Program 1:

Output:

37651

Explanation:

Here, we declared three local variables num, R1, and R2, and we are calculating the reverse of variable num, and R1 is used to extract the last digit in each iteration and R2 to store the result.

Iteration 1:
num=15673, R1=0, R2=0.
After executing loop statements R1=3, R2=3, and num=1567.

Iteration 2:
num=1567, R1=3, R2=3.
After executing loop statements R1=7, R2=37, and num=156.

Iteration 3:
num=156, R1=6, R2=37.
After executing loop statements R1=6, R2=376, and num=15.

Iteration 4:
num=15, R1=5, R2=376.
After executing loop statements R1=5, R2=3765, and num=1.

Iteration 5:
num=1, R1=1, R2=3765.
After executing loop statements R1=1, R2=37651 and num=0.

Then the condition will false and print "37651"

Answer Program 2:

Output:

0 2 6 12 20

Explanation:

In the above program, we declared three local variables I, D, and R.

Iteration 1:
I=1, D=0, R=0
R = 1*0
R = 0
Then I=2 and D=1 and loop condition is true. 

Iteration 2:
I=2, D=1, R=0
R = 2*1
R = 2
Then I=3 and D=2 and loop condition is true.

Iteration 3:
I=3, D=2, R=2
R = 3*2
R = 6
Then I=4 and D=3 and loop condition is true.

Iteration 4:
I=4, D=3, R=6
R = 4*3
R = 12
Then I=5 and D=4 and loop condition is true.

Iteration 5:
I=5, D=4, R=12
R = 5*4
R = 20
Then I=6 and D=5 and loop condition is false. 
And program terminates.

Answer Program 3:

Output:

1 9 16 25

Explanation:

In the above program, we declared three local variables I, D, and R.

Iteration 1:
I=1, D=1, R=0
R = 1 * 1
R = 1
Print the value of R that is 1.
Then I=2 and D=2 and loop condition is true. 

Iteration 2:
I=2, D=2, R=1
R = 2 * 2
R = 4
But it will skip "cout" statement because 
of the continue statement.
Then I=3 and D=3 and loop condition is true. 

Iteration 3:
I=3, D=3, R=4
R = 3 * 3
R = 9
Print the value of R that is 9.
Then I=4 and D=4 and loop condition is true. 

Iteration 4:
I=4, D=4, R=9
R = 4 * 4
R = 16
Print the value of R that is 16.
Then I=5 and D=5 and loop condition is true. 

Iteration 5:
I=5, D=5, R=16
R = 5 * 5
R = 25

Print the value of R that is 25.
Then I=6 and D=6 and loop condition is false. 
Then the loop will terminate.

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