According to the C++ Standard *8.5.1 Aggregates)
7 If there are fewer initializer-clauses in the list than there are
members in the aggregate, then each member not explicitly initialized
shall be initialized from its brace-or-equal-initializer or, if there
is no brace-or-equalinitializer, from an empty initializer list
(8.5.4).
and (8.5.4 List-initialization)
— Otherwise, if the initializer list has no elements, the object is
value-initialized.
and (8.5 Initializers)
— if T is an array type, then each element is value-initialized; —
otherwise, the object is zero-initialized.
Thus in both these definitions
char array[10][10] = {'/'}
int array[10][10] = {0}
all elements that do not have a corresponding initializer are value initialized that for fundamentals types means zero initialization. All elements except first elements of the arrays will be set to zero. That means that the both arrays from your example "work" the same way.:)
NULLcharacters?int array[10][10] = {1};? I would guess not.