How to print the array?

Question

int main() {     int my array[3][3] =     10, 23, 42,         1, 654, 0,       40652, 22, 0     };      printf("%d\n", my_array[3][3]);     return 0; } 

I am not able to get the array to print.. Any ideas why? I am a beginning programmer so any words of advice are appreciated.

Answer 1

What you are doing is printing the value in the array at spot [3][3], which is invalid for a 3by3 array, you need to loop over all the spots and print them.

for(int i = 0; i < 3; i++) {     for(int j = 0; j < 3; j++) {         printf("%d ", array[i][j]);     }     printf("\n"); }  

This will print it in the following format

10 23 42 1 654 0 40652 22 0 

if you want more exact formatting you'll have to change how the printf is formatted.

Answer 2

There is no .length property in C. The .length property can only be applied to arrays in object oriented programming (OOP) languages. The .length property is inherited from the object class; the class all other classes & objects inherit from in an OOP language. Also, one would use .length-1 to return the number of the last index in an array; using just the .length will return the total length of the array.

I would suggest something like this:

int index; int jdex; for( index = 0; index < (sizeof( my_array ) / sizeof( my_array[0] )); index++){    for( jdex = 0; jdex < (sizeof( my_array ) / sizeof( my_array[0] )); jdex++){         printf( "%d", my_array[index][jdex] );         printf( "\n" );    } } 

The line (sizeof( my_array ) / sizeof( my_array[0] )) will give you the size of the array in question. The sizeof property will return the length in bytes, so one must divide the total size of the array in bytes by how many bytes make up each element, each element takes up 4 bytes because each element is of type int, respectively. The array is of total size 16 bytes and each element is of 4 bytes so 16/4 yields 4 the total number of elements in your array because indexing starts at 0 and not 1.