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

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 sizeofoperator:

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 dividethe 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 ofa changed you would have a nasty bug if you forgot to changethe 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 parameterizethe array name in a macro and get:

#define NELEMS(x)  (sizeof(x) / sizeof((x)[0]))int a[17];size_t n = NELEMS(a);

The sizeof way is the right way iff you are dealing with arrays not received as parameters. An array sent as a parameter to a function is treated as a pointer, so sizeof will return the pointer's size, instead of the array's.

Thus, inside functions this method does not work. Instead, always pass an additional parameter size_t size indicating the number of elements in the array.

Test:

#include <stdio.h>#include <stdlib.h>void printSizeOf(int intArray[]);void printLength(int intArray[]);int main(int argc, char* argv[]){    int array[] = { 0, 1, 2, 3, 4, 5, 6 };    printf("sizeof of array: %d\n", (int) sizeof(array));    printSizeOf(array);    printf("Length of array: %d\n", (int)( sizeof(array) / sizeof(array[0]) ));    printLength(array);}void printSizeOf(int intArray[]){    printf("sizeof of parameter: %d\n", (int) sizeof(intArray));}void printLength(int intArray[]){    printf("Length of parameter: %d\n", (int)( sizeof(intArray) / sizeof(intArray[0]) ));}

Output (in a 64-bit Linux OS):

sizeof of array: 28sizeof of parameter: 8Length of array: 7Length of parameter: 2

Output (in a 32-bit windows OS):

sizeof of array: 28sizeof of parameter: 4Length of array: 7Length of parameter: 1

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));