CODEWITHSUNDEEP
cpointerslinked-list
Kevin
Kevin

Reputation: 437

Declaring a pointer type in the parameter of a function to modify a linked list

I know that in order for a function to write to a parameter, we need to pass a pointer to that parameter. That's also the case if we want to modify a linked list. The thing that I struggle to understand, is that why we need to declare a pointer type P* instead of declaring just the normal type in order to modify the linked list?

What are the main differences between the two examples?

int function(struct list_t** list){

}
int function(struct list_t* list){

}

I can say that in both the examples, I'm passing the pointer to the parameter, which technically allows me to edit the linked list.

Upvotes: 1

Views: 53

Answers (1)

Some programmer dude
Some programmer dude

Reputation: 409166

With struct list_t* list then list is a pointer to a list_t.

With struct list_t** list then list is a pointer to a pointer to a list_t.

If you have e.g.

list_t* my_list;

then my_list is a pointer to a list_t. If you then use the address-of operator & to get a pointer to my_list (as in &my_list) then you get a pointer to a pointer to a list_t, which is list_t**.

The use of pointers to pointers is to emulate pass-by-reference in C, which only have pass-by-value. As you say, if you have a pointer to a pointer, then you can dereference it and make it point somewhere else. With a plain pointer, it's passed by value and the value is copied into the argument variable. Modifying a copy will not modify the original.

Upvotes: 2

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