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Continue Statement in C/C++

Continue is also a loop control statement just like the break statement. continue statement is opposite to that of break statement, instead of terminating the loop, it forces to execute the next iteration of the loop.
As the name suggest the continue statement forces the loop to continue or execute the next iteration. When the continue statement is executed in the loop, the code inside the loop following the continue statement will be skipped and next iteration of the loop will begin.

Syntax
:

continue;


Example:
Consider the situation when you need to write a program which prints number from 1 to 10 and but not 6. It is specified that you have to do this using loop and only one loop is allowed to use.
Here comes the usage of continue statement. What we can do here is we can run a loop from 1 to 10 and every time we have to compare the value of iterator with 6. If it is equals to 6 we will use the continue statement to continue to next iteration without printing anything otherwise we will print the value.
Below is the implementation of above idea:

// C++ program to explain the use
// of continue statement
  
#include <iostream>
using namespace std;
  
int main()
{
    // loop from 1 to 10
    for (int i = 1; i <= 10; i++) {
  
        // If i is equals to 6,
        // continue to next iteration
        // without printing
        if (i == 6)
            continue;
  
        else
            // otherwise print the value of i
            cout << i << " ";
    }
  
    return 0;
}

Output:

1 2 3 4 5 7 8 9 10 

The continue statement can be used with any other loop also like while or do while in the similar way as it is used with for loop above.

Exercise Problem :
Given a number n, print triangular pattern. We are allowed to use only one loop.

Input: 7
Output:
*
* * 
* * *
* * * *
* * * * *
* * * * * *
* * * * * * *

Solution : Print the pattern by using one loop | Set 2 (Using Continue Statement)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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