How do i pass a function pointer to a function.

Question

Hey so i am trying to figure out how i pass a function pointer to a function stored within a struct. The following is the typedef

struct menu_item
{
    char name[ITEM_NAME_LEN+1];
    BOOLEAN (*func)(struct vm*);

};

The function i am trying to pass has the following prototype.

void print_list(struct vm_node  *root);

with the defintion of the file being:

void print_list(struct vm_node  *root) {
    while (root) {
        printf("%s",root->data->id);
        root = root->next;
    }
    printf("\n");
}

Krypton · Accepted Answer · 2014-10-07 03:43:22Z

1

struct menu_item item;
item.func = &print_list;

As simple as that. However BOOLEAN (*func)(struct vm*); needs to be changed to void (*func)(struct vm_node *); to match the destination function.

answered Oct 7, 2014 at 3:43

Krypton

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9 Comments

tangrs Over a year ago

The address of operator isn't needed at &print_list. print_list decays to a function pointer similar to how an array decays to a pointer.

Joshua Theeuf Over a year ago

@krypton the func will have to accept various other functions which will return other types.

Krypton Over a year ago

@tangrs: with ampersand is a more correct way of understanding.

Krypton Over a year ago

@JoshuaTheeuf: if you want it to accept various types function prototype, simply declare as void (*func)(), but then you would need to do type enforcing to prevent compiler warnings.

PM 2Ring Over a year ago

@JoshuaTheeuf the func will have to accept various other functions which will return other types - this is generally a bad idea. But if you really must do this, you can create a union of possible return types, and have all your functions return that union.

|

Farouq Jouti · Accepted Answer · 2014-10-07 03:44:01Z

0

functions are sort of like arrays so you can just do

menu_item.func = print_list;

just like you would do with an array

int arr[20];
int *ptr;
ptr = array /*

answered Oct 7, 2014 at 3:44

Farouq Jouti

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2 Comments

Krypton Over a year ago

Do you know what the above line does?

tangrs Over a year ago

OP wants a function pointer as far as I can see, not the result of a function call.

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