How to obtain sizes of arrays in a loop - c++

Question

I am aligning several arrays in order and performing some sort of classification. I created an array to hold other arrays in order to simplify the operations that I want to perform.
Sadly, my program crashed when I ran it and I went on to debug it to finally realize that the sizeof operator is giving me sizes of pointers and not arrays within the loop.So I resorted to the cumbersome solution and my program worked.
How can I avoid this cumbersome method? I want to calculate within a loop!
#include <iostream>
#include <string>
#define ARRSIZE(X) sizeof(X) / sizeof(*X)
int classify(const char *asset, const char ***T, size_t T_size, size_t *index);
int main(void)
{
const char *names[] = { "book","resources","vehicles","buildings" };
const char *books[] = { "A","B","C","D" };
const char *resources[] = { "E","F","G" };
const char *vehicles[] = { "H","I","J","K","L","M" };
const char *buildings[] = { "N","O","P","Q","R","S","T","U","V" };
const char **T[] = { books,resources,vehicles,buildings };
size_t T_size = sizeof(T) / sizeof(*T);
size_t n, *index = new size_t[T_size];
/* This will yeild the size of pointers not arrays...
for (n = 0; n < T_size; n++) {
index[n] = ARRSIZE(T[n]);
}
*/
/* Cumbersome solution */
index[0] = ARRSIZE(books);
index[1] = ARRSIZE(resources);
index[2] = ARRSIZE(vehicles);
index[3] = ARRSIZE(buildings);
const char asset[] = "L";
int i = classify(asset, T, T_size, index);
if (i < 0) {
printf("asset is alien !!!\n");
}
else {
printf("asset ---> %s\n", names[i]);
}
delete index;
return 0;
}
int classify(const char *asset, const char ***T, size_t T_size, size_t *index)
{
size_t x, y;
for (x = 0; x < T_size; x++) {
for (y = 0; y < index[x]; y++) {
if (strcmp(asset, T[x][y]) == 0) {
return x;
}
}
}
return -1;
}

Answer 1

As you are including <string> and <iostream> I assume that the question is about C++ and not C. To avoid all this complication, simply use containers. E.g:
#include <vector>
std::vector<int> vect = std::vector<int>(3,0);
std::cout << vect.size() << std::endl; // prints 3

Answer 2

One solution if you are coding in C is to terminate your array with a special item, like NULL
const char *books[] = { "A","B","C","D", NULL };
size_t size(const char *arr[])
{
const char **p = arr;
while (*p)
{
p++;
}
return p - arr;
}

Answer 3

You can specify the array size explizit:
size_t n, index[] = {ARRSIZE(books), ARRSIZE(resources), ARRSIZE(vehicles), ARRSIZE(vehicles)};
or if you want to avoid double typing you can you X-Macros to roll out everything:
#define TBL \
X(books) \
X(resources) \
X(vehicles) \
X(buildings)
const char **T[] = {
#define X(x) x,
TBL
};
#undef X
size_t n, index[] = {
#define X(x) ARRSIZE(x),
TBL
};
which produces the same. See Running Demo.