Passing an array by reference in C?

Question

How can I pass an array of structs by reference in C?

As an example:

struct Coordinate {    int X;    int Y; }; SomeMethod(Coordinate *Coordinates[]){    //Do Something with the array } int main(){     Coordinate Coordinates[10];    SomeMethod(&Coordinates); } 

Answer 1

In C arrays are passed as a pointer to the first element. They are the only element that is not really passed by value (the pointer is passed by value, but the array is not copied). That allows the called function to modify the contents.

void reset( int *array, int size) {    memset(array,0,size * sizeof(*array)); } int main() {    int array[10];    reset( array, 10 ); // sets all elements to 0 } 

Now, if what you want is changing the array itself (number of elements...) you cannot do it with stack or global arrays, only with dynamically allocated memory in the heap. In that case, if you want to change the pointer you must pass a pointer to it:

void resize( int **p, int size ) {    free( *p );    *p = malloc( size * sizeof(int) ); } int main() {    int *p = malloc( 10 * sizeof(int) );    resize( &p, 20 ); } 

In the question edit you ask specifically about passing an array of structs. You have two solutions there: declare a typedef, or make explicit that you are passing an struct:

struct Coordinate {    int x;    int y; }; void f( struct Coordinate coordinates[], int size ); typedef struct Coordinate Coordinate;  // generate a type alias 'Coordinate' that is equivalent to struct Coordinate void g( Coordinate coordinates[], int size ); // uses typedef'ed Coordinate 

You can typedef the type as you declare it (and it is a common idiom in C):

typedef struct Coordinate {    int x;    int y; } Coordinate;