Passing array as parameter - c

Question

I want to pass an array as a parameter to another function:
int i;
int main()
{
int a[5] = {1, 2, 3, 4, 5};
printf("Main:\n");
for(i=0; i<sizeof(a)/sizeof(int); i++)
{
printf("%i\n", a[i]);
}
func(a);
return;
}
void func(int a[])
{
printf("Func:\n");
for(i=0; i<sizeof(a)/sizeof(int); i++)
{
printf("%i\n", a[i]);
}
}
The loop in the main function prints all 5 values:
Main:
1
2
3
4
5
But the loop in the function func only prints 2 values:
Func:
1
2
Why this strange behaviour?

Answer 1

I want to pass an array as a parameter to another function:
This is a common pitfall. Arrays do not bring along their length, as they do in other languages. A C "array" is just a bunch of contiguous values, so sizeof will not (necessarily) return the length of the array.
What actually happens is that the function gets passed a pointer to the area of memory where the array is stored (and therefore, to the first element of the array), but no information about that area's size. To "pass an array with size", you must do something to provide the extra information:
explicitly pass also its length as an extra parameter. Safer: you can pass uninitialized arrays.
use a special "terminating" value on the array. More compact: you pass only one parameter, the pointer to the array.
(suggested implicitly by #CisNOTthatGOODbutISOisTHATBAD's comment): pass a pointer to a struct holding a pointer to the memory and a size_t length in elements (or in bytes). This has the advantage of allowing to store yet more metadata about the array.
For arrays of integral type, you could even store the length in the first (zeroth) element of the array. This can sometimes be useful when porting from languages that have 'measured' arrays and indexes starting from 1. In all other cases, go for method #1. Faster, safer, and in my opinion clearer.
Strings are arrays of characters that employ a variation of method #2: they are terminated by a special value (zero).
Using method #1, your function would become:
void func(size_t n, int a[])
{
printf("Func:\n");
for (i=0; i < n; i++)
{
printf("%i\n", a[i]);
}
}
(it is equivalent to void func(size_t n, int *a) ).

Answer 2

Declaring a function with a parameter of array type is equivalent to declaring it with a parameter of pointer type, i.e. your function is equivalent to:
void func(int *a)
As such, the computation of sizeof(a) inside func computes the size of an int *, not the size of the original array (5 * sizeof(int)).
Since the size of an int * on your platform is apparently twice the size of an int, you see two values printed inside the function in contrast to the five printed outside it (where sizeof(a) correctly computes the size of the array).
This is all related to the fact that when you pass an array to a function, what you're actually doing is passing a pointer to its first element.
Note in passing that this is a FAQ of the comp.lang.c newsgroup:
http://c-faq.com/~scs/cgi-bin/faqcat.cgi?sec=aryptr#aryparmsize

Answer 3

I would change the code to this:
int i;
const int N = 5;
int main()
{
int a[N] = {1, 2, 3, 4, 5};
printf("Main:\n");
for(i=0; i<N; i++)
{
printf("%i\n", a[i]);
}
func(a);
return;
}
void func(int a[])
{
printf("Func:\n");
for(i=0; i<N; i++)
{
printf("%i\n", a[i]);
}
}
Parameter N is constant and define size of array.
I think it is better way, then 2 parameters in function.