I need to know if i want to make an array that every element of the array is a pointer to a linked list and pass the array to a function, the function is void because I need to change the array
typedef struct n{
char *S;
int num;
}list;
int (main){
list *Array[50];
return 0;
}
should the function be void changeArray(list A[]); or void changeArray(list *A[]); or void changeArray(list **A[]);
The function could be either void changeArray(list *A[]) or void changeArray(list **A). Both signatures would accept an array of pointers, and let you change elements of that array:
void changeArray(list *A[]) {
...
A[0] = malloc(list);
}
A inside your function, the array that you pass will be modified. If you would like to pass a copy of the array, either wrap it in a struct, or make a copy explicitly with memcpy.void change(list A[]); Array is an array of pointers. Each element of the array is a pointer to list, but the array itself becomes a two-star-pointer, i.e. a pointer to a pointer. Had Array been declared as list Array[50], then void change(list A[]); would be the right signature. However, since you declared it with an asterisk, i.e. list *Array[50], you need a second asterisk in the signature of the function that receives your array.The array is defined like
list *Array[50];
So if you want to pass this array to a function then it should be declared like
void changeArray( list *Array[50], size_t n );
or
void changeArray( list *Array[], size_t n );
or
void changeArray( list **Array, size_t n );
and called like
changeArray( Array, 50 );
In any case a declaration of an array as a function parameter is adjusted to a pointer to object of the type of elements of the array.
void changeArray(list *A[]);