Executive summary:
int a[17];
size_t n = sizeof(a)/sizeof(a[0]);
Full answer:
To determine the size of your array in bytes, you can use the sizeof
operator:
int a[17];
size_t n = sizeof(a);
On my computer, ints are 4 bytes long, so n is 68.
To determine the number of elements in the array, we can divide
the total size of the array by the size of the array element.
You could do this with the type, like this:
int a[17];
size_t n = sizeof(a) / sizeof(int);
and get the proper answer (68 / 4 = 17), but if the type of
a changed you would have a nasty bug if you forgot to change
the sizeof(int) as well.
So the preferred divisor is sizeof(a[0]) or the equivalent sizeof(*a), the size of the first element of the array.
int a[17];
size_t n = sizeof(a) / sizeof(a[0]);
Another advantage is that you can now easily parameterize
the array name in a macro and get:
#define NELEMS(x) (sizeof(x) / sizeof((x)[0]))
int a[17];
size_t n = NELEMS(a);
Unless that value is 0 (in which case you can omit some part of the initializer
and the corresponding elements will be initialized to 0), there's no easy way.
Don't overlook the obvious solution, though:
int myArray[10] = { 5, 5, 5, 5, 5, 5, 5, 5, 5, 5 };
Elements with missing values will be initialized to 0:
int myArray[10] = { 1, 2 }; // initialize to 1,2,0,0,0...
So this will initialize all elements to 0:
int myArray[10] = { 0 }; // all elements 0
In C++, an empty initialization list will also initialize every element to 0.
This is not allowed with C:
int myArray[10] = {}; // all elements 0 in C++
Remember that objects with static storage duration will initialize to 0 if no
initializer is specified:
static int myArray[10]; // all elements 0
And that "0" doesn't necessarily mean "all-bits-zero", so using the above is
better and more portable than memset(). (Floating point values will be
initialized to +0, pointers to null value, etc.)
Best Answer
Here's code to get the full path to the executing app:
Windows:
Linux: