FAQ: Arrays: Lesson - Element Access in Multidimensional Arrays

This community-built FAQ covers the “Element Access in Multidimensional Arrays” exercise from the lesson “Arrays: Lesson”.

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Why does the lesson require using sizeof(double) when the matrix is made of ints? And does sizeof(matrix[0][0]) work as well?

5 Likes

Same question here. Could it be a mistake?
The array matrix was initialized as an int array.
When using sizeof(double) in the nested for loop to determine the length of the columns, not all numbers get printed out.
Here is my solution (I added print statements to check if my loops iterate through all numbers in the array and they do.):

#include<stdio.h>

int main() {
  int matrix[][4] = {{14, 10, 6, 4}, {3, 7, 18, 11}, {13, 9, 5, 17}, {19, 12, 2, 1}}; 
  int sum = 0;

  // Checkpoint 1 code goes here.
  printf("%i\n", matrix[3][1]);
  // Checkpoint 2 code goes here.
  for (int i = 0; i < sizeof(matrix)/sizeof(matrix[0]); i++) {
    for (int j = 0; j < sizeof(matrix[0])/sizeof(int); j++) {
      printf("%i\t", matrix[i][j]);
      sum += matrix[i][j];
    }
  }
  printf("%i\n", sum);
}
1 Like

Hey @toshymoshy , thank you for u explanation, I wish to have another one…

can u explain why in the inner loop u mention sizeof(int) ? r u intentionally using the size of the datatype ?

EDIT: after reading this everything fell into place: How do I determine the size of my array in C? - Stack Overflow

the middle row in the snippet:

  for (int i = 0; i < sizeof(matrix)/sizeof(matrix[0]); i++) {
    for (int j = 0; j < sizeof(matrix[0])/sizeof(int); j++) {  // <<< here
      printf("%i\t", matrix[i][j]);

Thanks in advance :slight_smile:

1 Like