C++ Continue in for loop

Continue statement:

Inside the loop, the continue statement is used to control the loop.

C++ utilizes the continue keyword to implement the continue statement, which transfers the program's flow at the start of the loop and skips the current statement when it is reached. Continue statement is used in all the iteration control loops.

Syntax:

The syntax of the continue statement is: 
         continue;

• Continue is a predefined keyword.
• The continue statement can’t be used outside of the iteration loop. It only appears in iteration statements.

Flowchart:

Explanation:

The statements within the loop are executed based on the loop condition. If the specified condition is true, the loop iteration will be continued to execute and the loop is terminated when the specified condition is false. The continue statement skips the current statement and passes control to the next loop iteration when the continue keyword is used inside a loop.

What is For Loop?

A for loop is a repetitive control structure that allows you to create a loop to execute a specific number of times efficiently.

The syntax that can be used for the for loop in the C++ programming language is given below:

for (initialization statement; termination condition; increment or decrement statement (used for modifying value for further evaluation))
{
    /* main body of the “for” loop */
}

Continue in For Loop

The continue keyword in the for loop moves the control to the update statement of the loop in which it is used.

  for ( , , ) {         
   {                                           
     // ...            
  continue;
   }
 } 

Examples of Continue in Do-While loop

Algorithm to print the list of numbers:

Step 1: Start the program.

Step 2: Read n and i.

Step 3: Initialise i=1.

Step 4: Print the value of I,

If (i==3)

Continue;

Else

go to step 5.

Step 5: Iterate until I<=10 if the condition is false then go to step 4.

Step 6: Stop.

Program:

#include<iostream>
using namespace std;
int main()
{
int i, n;
cout << "Enter a number";
cin >> n;
cout << "the list of numbers are";
for (i=1; i<=n; i++)
{
   if(i == 5)
    {
      continue;
   }
 cout << i;
  }
return 0;
}

The output of the program

Enter a number: 10
The numbers are:
1
2
3
4
6
7
8
9
10

Explanation

The program is designed to print the list of numbers which are entered by the user.

We have used the for loop method in the above code to demonstrate the implementation of the continue statement.

We've set up two variables: "n" for entering the value and "i" for the iteration. The program asks the user for a number, which is then stored in variable ‘n’. We initialized i to 1(i=1) and used a condition as text expression that verifies the value of "i" is less than or equal to n (i<=n). In the loop, we used the continue statement in the if condition to skip number five. If the number is not 5, the statement outside the ‘if condition’ is executed, and the number is printed one by one.

Algorithm to print Odd numbers using continue statement:

Step 1: Start the program.

Step 2: Read n and i.

Step 3: Initialise i=1.

Step 4: Iterate until i<=10 if the condition is false then go to step 6.

Step 5: Print the value of I.

If (i%2==0)

Continue;

Else

go to step 4.

Step 6: exit

Program:

#include<iostream>
using namespace std;
int main()
{
int i, n;
cout << "Enter a number";
cin >> n;
 cout << "Odd numbers";
 for (i=1; i<=n; i++)
{
   if(i%2==0)
    {
      continue;
   }
 cout<<i;
    }
return 0;
}

Output:

Enter a number: 10
Odd numbers:
1
3
5
7
9

Explanation

We have used the for loop method in the above code to demonstrate the implementation of the continue statement.

We've set up two variables: "n" for entering the value and "i" for the iteration. The program asks the user for a number, which is then stored in variable ‘n’. We initialized i to 1(i=1) and used a condition as text expression that verifies the value of "i" is less than or equal to n (i<=n).

After then, we used the continue statement inside the loop to skip the even numbers with the logic (i%2==0). if the condition is true according to the logic then the number is skipped. And, if the number is not even, the statement outside if condition is executed, and the odd numbers are printed one by one.

Continue with the Nested loop:

When the continue keyword is used with nested loops, the current iteration of the inner loop is skipped.

Program:

// using continue statement inside
// nested for loop
#include <iostream>
using namespace std;
int main()
 {
    int n;
    int sum=0;
// nested for loops
            // first loop
  for (int i = 1; i <= 5; i++) 
{
        // second loop
     for (int j = 1; j <= 5; j++)
 {
            if (j ==3) 
             {
                continue;
            }
            cout << "i = " << i << ", j = " << j << endl;
  }
 }
return 0;
}

Output

i = 1, j = 2
i = 1, j = 4
i = 1, j = 5
i = 2, j = 2
i = 2, j = 4
i = 2, j = 5
i = 3, j = 2
i = 3, j = 4
i = 3, j = 5
i = 4, j = 2
i = 4, j = 4
i = 4, j = 5

Explanation

When the continue statement is used in the above program, it skips the current iteration of the inner loop. And the program's control shifts to the update expression of inner loop. As a result, the value of j=3 never appears in the output.