CODEWITHSUNDEEP
c++pointers
Anas
Anas

Reputation: 379

Passing the address of a pointer as a parameter

What is the point passing the address of a pointer as a parameter? For example:

int *ptr_one;
ptr_one = (int *)malloc(sizeof(int));

then function is being called as the following:

func(&ptr_one);

Note the function argument is as follow:

func(int **ptr)
{
......
}

Upvotes: 1

Views: 504

Answers (2)

wolfPack88
wolfPack88

Reputation: 4203

If you wish to modify the pointer itself, then you need to pass it either by reference, or by the address of the pointer. E.g.,

func(int **ptr)
{
    free(ptr);

    ptr = new int[2]; //side note: use new instead of malloc in C++
                      //memory allocated with new is deallocated with
                      //delete
}

In C, passing the address of a pointer was the only way you could allocate or deallocate the memory of that pointer in a function. In C++, however, it is usually better to pass by reference instead. So a C++ version of your code would look like:

int *ptr = new int;
func(ptr);
delete ptr;

void func(int *&ptr) {
.
.
.
}

Upvotes: 2

Thomas Matthews
Thomas Matthews

Reputation: 57749

The purpose of passing a pointer to a pointer is so that the pointer variable can be modified.

Recall, from the C-style of coding, that a parameter can be modified by passing the address or pointer to the parameter. If the parameter is an int that needs to be modified, a pointer to the integer is passed. Likewise, if a pointer parameter will be modified by the function, it was passed by pointer to pointer (or address of pointer).

Upvotes: 2

Related Questions