Why is the arr[i][j-1] zero? After testing using the printf function, it shows that the position is arr[0][-1] and it has the value 0 as it appears from the results below.
int main() {
int arr [2][4];
int i,j;
for (i=0; i<2; i++) {
for (j = 0; j < 4; j++) {
printf("Provide a number: ");
scanf("%d", &arr[i][j]); // get the values
printf("The arr[i][j] is %d\n",arr[i][j]);
printf("The arr[i][j-1] is %d\n",arr[i][j-1]);
}
}
//print the values
for (i=0; i<2; i++) {
for (j = 0; j < 4; j++) {
printf("%d\n",arr[i][j]);
}
}
}
Example of how it works:
Provide a number: 12
The arr[i][j] is 12
The arr[i][j-1] is 0
Provide a number: 14
The arr[i][j] is 14
The arr[i][j-1] is 12
Provide a number:
If the following arr[i][j-2] is used though, then it shows this:
The arr[i][j-2] is 6611520
this is typical in C but the 0 in j-1 is not clear. Can you explain?
>Solution :
arr[0][-1] is outside the bounds of the array.
Reading or writing outside the bounds of an array (or even creating a pointer to such an element) triggers undefined behavior. Loosely speaking, this means there are no guarantees regarding what your program will do.