C 函数返回指针和动态数组

Question

I have a function which creates a dynamic array the size of my selected integer. Code:

int *create(int n) {
    int *nn;
    nn = (int*)malloc(n*sizeof(int));
    return nn;
}

And I call it like this in my main()

int *nn
int n = 5; /* size = 5 */
nn = create(n);

I think I get the part, int *create(...) which is supposed to return the address to the first position of my returned nn. However I wonder, is there a way to not use the pointer in a function and instead modify the return nn; part so that I still return the address of the dynamic array nn ?

I want to remove the * in the *create

Answer 1

You can modify your code to set a pointer passed into your function by address, like this:

void create(int n, int** res) {
    *res = malloc(n*sizeof(int));
}

Here is how you call this function now:

int *nn
int n = 5;
create(n, &nn);

I want to use only one parameter and be able to do it with a return .

You can return the result of malloc directly, since that is all your function does anyway:

int *create(int n) {
    return malloc(n*sizeof(int));
}

The call can be combined with the declaration as well:

int n = 5;
int *nn = create(n);

Of course if you do that, you might as well do the idiomatic

int *nn = malloc(n*sizeof(*nn));

and drop the create function altogether.

Answer 2

You could do :

int *create(int n) 
{
    return malloc(n*sizeof(int));
}

This way you don't need a temporary variable inside your create() function. But I don't see the point to make a function that does only a call to malloc .

If you want your function to return an array of integer, you need your function to be in the form int *create() since an array can be considered a pointer to its first element, that is an integer in your case.