Nested Loops in C++ with Examples

Last Updated : 12 Jul, 2025

Nested loop means a loop statement inside another loop statement. That is why nested loops are also called as "loop inside loop". Syntax for Nested For loop:

 for ( initialization; condition; increment ) { for ( initialization; condition; increment ) { // statement of inside loop } // statement of outer loop }

Syntax for Nested While loop:

 while(condition) { while(condition) { // statement of inside loop } // statement of outer loop }

Syntax for Nested Do-While loop:

 do{ do{ // statement of inside loop }while(condition); // statement of outer loop }while(condition);

Note: There is no rule that a loop must be nested inside its own type. In fact, there can be any type of loop nested inside any type and to any level.

Syntax:

 do{ while(condition) { for ( initialization; condition; increment ) { // statement of inside for loop } // statement of inside while loop } // statement of outer do-while loop }while(condition);

Below are some examples to demonstrate the use of Nested Loops: Example 1: Below program uses a nested for loop to print a 2D matrix of 3x3.

CPP

// C++ program that uses nested for loop // to print a 2D matrix #include <bits/stdc++.h> using namespace std; #define ROW 3 #define COL 3 // Driver program int main() {  int i, j;  // Declare the matrix  int matrix[ROW][COL] = { { 1, 2, 3 },  { 4, 5, 6 },  { 7, 8, 9 } };  cout << "Given matrix is \n";  // Print the matrix using nested loops  for (i = 0; i < ROW; i++) {  for (j = 0; j < COL; j++)  cout << matrix[i][j];  cout << "\n";  }  return 0; }

Output:

 Given matrix is 123 456 789

Example 2: Below program uses a nested for loop to print all prime factors of a number.

CPP

// C++ Program to print all prime factors // of a number using nested loop #include <bits/stdc++.h> using namespace std; // A function to print all prime factors of a given number n void primeFactors(int n) {  // Print the number of 2s that divide n  while (n % 2 == 0) {  cout << 2;  n = n / 2;  }  // n must be odd at this point. So we can skip  // one element (Note i = i +2)  for (int i = 3; i <= sqrt(n); i = i + 2) {  // While i divides n, print i and divide n  while (n % i == 0) {  cout << i;  n = n / i;  }  }  // This condition is to handle the case when n  // is a prime number greater than 2  if (n > 2)  cout << n; } /* Driver program to test above function */ int main() {  int n = 315;  primeFactors(n);  return 0; }