Jagged Array or Array of Arrays in C with Examples

14 Feb, 2020
Prerequisite: Arrays in C
Jagged array is array of arrays such that member arrays can be of different sizes, i.e., we can create a 2-D array but with a variable number of columns in each row. These type of arrays are also known as Jagged arrays.
Example:
arr[][] = { {0, 1, 2},
            {6, 4},
            {1, 7, 6, 8, 9},
            {5} 
          };

Below are the methods to implement the jagged array in C:
Using array and a pointer (Static Jagged Array)First declare 1-D arrays with the number of rows you will need,The size of each array (array for the elements in the row) will be the number of columns (or elements) in the row,Then declare a 1-D array of pointers that will hold the addresses of the rows,The size of the 1-D array is the number of rows you want in the jagged array.Below is the implementation of the above approach:Example:// C program to show the// implementation of Jagged Arrays  #include <stdio.h>#include <stdlib.h>  int main(){      int row0[4] = { 1, 2, 3, 4 };    int row1[2] = { 5, 6 };      int* jagged[2] = { row0, row1 };      // Array to hold the size of each row    int Size[2] = { 4, 2 }, k = 0;      // To display elements of Jagged array    for (int i = 0; i < 2; i++) {          // pointer to hold the address of the row        int* ptr = jagged[i];          for (int j = 0; j < Size[k]; j++) {            printf("%d ", *ptr);              // move the pointer to the            // next element in the row            ptr++;        }          printf("\n");        k++;          // move the pointer to the next row        jagged[i]++;    }      return 0;}Output:1 2 3 4 
5 6
Using an array of pointer (Dynamic Jagged Array)Declare an array of pointers (jagged array),The size of this array will be the number of rows required in the Jagged arrayThen for each pointer in the array allocate memory for the number of elements you want in this row.Below is the implementation of the above approach:Example:// C program to show the// implementation of Jagged Arrays  #include <stdio.h>#include <stdlib.h>  int main(){    // 2 Rows    int* jagged[2];      // Allocate memory for elements in row 0    jagged[0] = malloc(sizeof(int) * 1);      // Allocate memory for elements in row 1    jagged[1] = malloc(sizeof(int) * 3);      // Array to hold the size of each row    int Size[2] = { 1, 3 }, k = 0, number = 100;      // User enters the numbers    for (int i = 0; i < 2; i++) {          int* p = jagged[i];          for (int j = 0; j < Size[k]; j++) {            *p = number++;              // move the pointer            p++;        }        k++;    }      k = 0;      // Display elements in Jagged array    for (int i = 0; i < 2; i++) {          int* p = jagged[i];        for (int j = 0; j < Size[k]; j++) {              printf("%d ", *p);            // move the pointer to the next element            p++;        }        printf("\n");        k++;        // move the pointer to the next row        jagged[i]++;    }      return 0;}Output:100 
101 102 103

Using array and a pointer (Static Jagged Array)First declare 1-D arrays with the number of rows you will need,The size of each array (array for the elements in the row) will be the number of columns (or elements) in the row,Then declare a 1-D array of pointers that will hold the addresses of the rows,The size of the 1-D array is the number of rows you want in the jagged array.Below is the implementation of the above approach:Example:// C program to show the// implementation of Jagged Arrays  #include <stdio.h>#include <stdlib.h>  int main(){      int row0[4] = { 1, 2, 3, 4 };    int row1[2] = { 5, 6 };      int* jagged[2] = { row0, row1 };      // Array to hold the size of each row    int Size[2] = { 4, 2 }, k = 0;      // To display elements of Jagged array    for (int i = 0; i < 2; i++) {          // pointer to hold the address of the row        int* ptr = jagged[i];          for (int j = 0; j < Size[k]; j++) {            printf("%d ", *ptr);              // move the pointer to the            // next element in the row            ptr++;        }          printf("\n");        k++;          // move the pointer to the next row        jagged[i]++;    }      return 0;}Output:1 2 3 4 
5 6

First declare 1-D arrays with the number of rows you will need,
The size of each array (array for the elements in the row) will be the number of columns (or elements) in the row,
Then declare a 1-D array of pointers that will hold the addresses of the rows,
The size of the 1-D array is the number of rows you want in the jagged array.
Below is the implementation of the above approach:Example:
// C program to show the// implementation of Jagged Arrays  #include <stdio.h>#include <stdlib.h>  int main(){      int row0[4] = { 1, 2, 3, 4 };    int row1[2] = { 5, 6 };      int* jagged[2] = { row0, row1 };      // Array to hold the size of each row    int Size[2] = { 4, 2 }, k = 0;      // To display elements of Jagged array    for (int i = 0; i < 2; i++) {          // pointer to hold the address of the row        int* ptr = jagged[i];          for (int j = 0; j < Size[k]; j++) {            printf("%d ", *ptr);              // move the pointer to the            // next element in the row            ptr++;        }          printf("\n");        k++;          // move the pointer to the next row        jagged[i]++;    }      return 0;}
1 2 3 4 
5 6

Using an array of pointer (Dynamic Jagged Array)Declare an array of pointers (jagged array),The size of this array will be the number of rows required in the Jagged arrayThen for each pointer in the array allocate memory for the number of elements you want in this row.Below is the implementation of the above approach:Example:// C program to show the// implementation of Jagged Arrays  #include <stdio.h>#include <stdlib.h>  int main(){    // 2 Rows    int* jagged[2];      // Allocate memory for elements in row 0    jagged[0] = malloc(sizeof(int) * 1);      // Allocate memory for elements in row 1    jagged[1] = malloc(sizeof(int) * 3);      // Array to hold the size of each row    int Size[2] = { 1, 3 }, k = 0, number = 100;      // User enters the numbers    for (int i = 0; i < 2; i++) {          int* p = jagged[i];          for (int j = 0; j < Size[k]; j++) {            *p = number++;              // move the pointer            p++;        }        k++;    }      k = 0;      // Display elements in Jagged array    for (int i = 0; i < 2; i++) {          int* p = jagged[i];        for (int j = 0; j < Size[k]; j++) {              printf("%d ", *p);            // move the pointer to the next element            p++;        }        printf("\n");        k++;        // move the pointer to the next row        jagged[i]++;    }      return 0;}Output:100 
101 102 103

Declare an array of pointers (jagged array),
The size of this array will be the number of rows required in the Jagged array
Then for each pointer in the array allocate memory for the number of elements you want in this row.
Below is the implementation of the above approach:Example:
// C program to show the// implementation of Jagged Arrays  #include <stdio.h>#include <stdlib.h>  int main(){    // 2 Rows    int* jagged[2];      // Allocate memory for elements in row 0    jagged[0] = malloc(sizeof(int) * 1);      // Allocate memory for elements in row 1    jagged[1] = malloc(sizeof(int) * 3);      // Array to hold the size of each row    int Size[2] = { 1, 3 }, k = 0, number = 100;      // User enters the numbers    for (int i = 0; i < 2; i++) {          int* p = jagged[i];          for (int j = 0; j < Size[k]; j++) {            *p = number++;              // move the pointer            p++;        }        k++;    }      k = 0;      // Display elements in Jagged array    for (int i = 0; i < 2; i++) {          int* p = jagged[i];        for (int j = 0; j < Size[k]; j++) {              printf("%d ", *p);            // move the pointer to the next element            p++;        }        printf("\n");        k++;        // move the pointer to the next row        jagged[i]++;    }      return 0;}
100 
101 102 103

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