How do I determine the size of my array in C?

Question

How do I determine the size of my array in C?

That is, the number of elements the array can hold?

Answer 1

If you know the data type of the array, you can use something like:

int arr[] = {23, 12, 423, 43, 21, 43, 65, 76, 22};

int noofele = sizeof(arr)/sizeof(int);

Or if you don't know the data type of array, you can use something like:

noofele = sizeof(arr)/sizeof(arr[0]);

Note: This thing only works if the array is not defined at run time (like malloc) and the array is not passed in a function. In both cases, arr (array name) is a pointer.

Answer 2

For multidimensional arrays it is a tad more complicated. Oftenly people define explicit macro constants, i.e.

#define g_rgDialogRows   2
#define g_rgDialogCols   7

static char const* g_rgDialog[g_rgDialogRows][g_rgDialogCols] =
{
    { " ",  " ",    " ",    " 494", " 210", " Generic Sample Dialog", " " },
    { " 1", " 330", " 174", " 88",  " ",    " OK",        " " },
};

But these constants can be evaluated at compile-time too with sizeof:

#define rows_of_array(name)       \
    (sizeof(name   ) / sizeof(name[0][0]) / columns_of_array(name))
#define columns_of_array(name)    \
    (sizeof(name[0]) / sizeof(name[0][0]))

static char* g_rgDialog[][7] = { /* ... */ };

assert(   rows_of_array(g_rgDialog) == 2);
assert(columns_of_array(g_rgDialog) == 7);

Note that this code works in C and C++. For arrays with more than two dimensions use

sizeof(name[0][0][0])
sizeof(name[0][0][0][0])

etc., ad infinitum.

Answer 3

#define SIZE_OF_ARRAY(_array) (sizeof(_array) / sizeof(_array[0]))

Answer 4

You can use the & operator. Here is the source code:

#include<stdio.h>
#include<stdlib.h>
int main(){

    int a[10];

    int *p; 

    printf("%p\n", (void *)a); 
    printf("%p\n", (void *)(&a+1));
    printf("---- diff----\n");
    printf("%zu\n", sizeof(a[0]));
    printf("The size of array a is %zu\n", ((char *)(&a+1)-(char *)a)/(sizeof(a[0])));


    return 0;
};

Here is the sample output

1549216672
1549216712
---- diff----
4
The size of array a is 10

Answer 5

It is worth noting that sizeof doesn't help when dealing with an array value that has decayed to a pointer: even though it points to the start of an array, to the compiler it is the same as a pointer to a single element of that array. A pointer does not "remember" anything else about the array that was used to initialize it.

int a[10];
int* p = a;

assert(sizeof(a) / sizeof(a[0]) == 10);
assert(sizeof(p) == sizeof(int*));
assert(sizeof(*p) == sizeof(int));

Answer 6

int a[10];
int size = (*(&a+1)-a) ;

for more details: https://aticleworld.com/how-to-find-sizeof-array-in-cc-without-using-sizeof/ https://www.geeksforgeeks.org/how-to-find-size-of-array-in-cc-without-using-sizeof-operator/