Created
March 5, 2020 12:44
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Hackerrank Dynamic arrays in C Demo
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#include<stdio.h> | |
int main() | |
{ | |
int *bookShelf[3]; // Array of pointer to int; | |
int shelf1[3] = {1, 2, 3}; // constains 3 books of total pages 1, 2 & 3. | |
int shelf2[4] = {4, 5, 6, 7}; | |
int shelf3[5] = {8, 9, 10, 11, 12}; | |
int shelvesSizes[3] = {3, 4, 5}; // It is stores no. of shelves | |
bookShelf[0] = shelf1; // base address of array is constant and row1 == & row1[0] | |
bookShelf[1] = &shelf2[0]; | |
bookShelf[2] = shelf3; // or &row[0]; | |
printf("Book Shelf:\n\nShelf\tBook Pages\n"); // Shelf no : i + 1 message | |
for(int i = 0 ; i < 3 ; i++) | |
{ | |
printf("%d : \t", i + 1); | |
for(int j = 0; j < shelvesSizes[i] ; j++) | |
printf("%d ", *(bookShelf[i] + j)); | |
printf("\n"); | |
} | |
return 0; | |
} |
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