Language C has not function available that recover size of array by default. However, we can use a trick that do possible get such size. Look like on example below:

1
2
3
4
5
6
7
8
9

#include <stdio.h>

int
main()
{
        int a[ 99 ];
        size_t n = sizeof( a ) / sizeof( int );
        printf( "%d\n", n );
}

All right, the output will be 99. Thus, if you change the array to char type, for example, you can have problems because the expression above use int type.

As see above, we know that, the C language use the unary operator function sizeof, that compute the size of type(char, int, float, double…) in compile time. So, the int type has a different size of char type. Also, the result of expression can be different relative from machine architecture. eg: 32 bits or 64 bits.

Then, to avoid this, we should use something like this way:

1
2
3
4
5
6
7
8
9

#include <stdio.h>

int
main()
{
        int a[ 99 ];
        size_t n = sizeof( a ) / sizeof( a[ 0 ] ); // Here, we get the type of array
        printf( "%d\n", n );
}

In example above, we get the first field of the array(sizeof(a[0])), instead to compute by type(sizeof(int)). It was sufficient to compute array size, independent of data type her. Notice also that, this became more generic expression.