How do you pass an array to a function in C programming?

Answer 1

The declaration of an array of 3 int pointers will be:

int *p[3];

The declaration of a function that receive an array of int pointers will be:

int my_func(int **p);

or

int my_func(int *p[]);

and you can pass it to other function simply by passing p.

Answer 2

Arrays are passed as pointers, eg:

void function (int *ip);

int main (void)

{

int n[32];

int *ip = &n[0];

/* these three do the same thing: */

function (&n[0]);

function (n);

function (ip);

}

Answer 3

You don't pass the entire array. You pass a single value, the address of one of its elements, usually the first element.

int a[] = {1, 3, 5, 7, 9, 2, 4, 6, 8, 10};

This creates an int array of length 10, with initial values as stated. The first element is a[0], having value of 1, and the last element is a[9], having value of 10.

somefunction (int * b);

somefunction(a);

This declares a function that receives an address to an int array. The call passes the address of a[0] to the function. You could just as easily pass some other element, such as a[3], by saying somefunction (&a[3]); or somefunction (a+3); Of course, in this case, you would not want to access b[9], because that would violate the addressability limit of the pointer, but you could certainly access b[3], which would be the same as a[6].

Note, in this example, that I mix array syntax with pointer syntax. In fact, the two are the same, and you can easily inter-mix them. You could just as easily say...

somefunction (int b[]);

somefunction(a);

Answer 4

eg:

void foo (char par [])

{

...

}

void bar (void)

{

char var [33];

foo (var); /* or: foo (&var[0]); */

}

Answer 5

type *A[] AS AN ARGUMENT