Example A: length is some integer variable
int values[length][2];
This array will not have all the elements inside as zeros.
Example B:
int values[10][2];
This one will be all set to zeros.
My main question is how can I make the first one be initialized as zeros?
But, I would like if someone could explain why is the behavior different?
Any variable defined in block scope (i.e. inside of a function or an enclosing block) is not implicitly initialized. It's values are indeterminate. This is the case with both of your examples. Just because the values happened to be 0 doesn't mean they were initialized.
If the array is question is not a variable length array, you could initialize it like this:
int values[10][2] = {{ 0 }};
This explicitly initializes one element of the 2D array to 0, and implicitly initializes the rest to 0.
A VLA however cannot be explicitly initialized. So you should use memset to set the bytes to 0.
int values[length][2];
memset(values, 0, sizeof values);
sizeof values here? Unknown at compile time.
sizeof operator is evaluated at run time instead of compile time.
memset does exactly what I was asking fortry using calloc it intialises junk of memory to 0's
the thing is, it might be initialized by 0's if you are lucky and depending on the compiler you use. The easiest might be:
memset(values, '\0', sizeof(int) * length * 2);
int values[10][2] = { 0 };but not the first. You must usememsetor loops.