C Function returning pointer and a dynamic array

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

Also, the canonical way to do this would just be int * nn = malloc(n*sizeof(int)); You don't cast malloc in c. So the example is a little confusing. — QuestionC
– QuestionC, Commented Oct 14, 2015 at 18:01
What is your reason for wanting eliminating the *? It's possible but it ain't pretty. — QuestionC
– QuestionC, Commented Oct 14, 2015 at 18:11
Why do you want to remove the * from the create function declaration/definition? What is your justification? Do you mean that you don't want the function to return a pointer value? If not, why not? — John Bode
– John Bode, Commented Oct 14, 2015 at 18:11

Sergey Kalinichenko · Accepted Answer · 2015-10-14 18:05:50Z

5

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.

edited Oct 14, 2015 at 18:05

answered Oct 14, 2015 at 18:01

Sergey Kalinichenko

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3 Comments

Rywi Over a year ago

Yes, that's one of the ways it's possible to solve it, however I want to use only one parameter and be able to do it with a return. Is it impossible to do so?

Weather Vane Over a year ago

Isn't this over-egging the pudding? See comment above from @QuestionC.

John Bode Over a year ago

"...and drop the create function altogether" is the right answer in this case, since create isn't doing anything other than calling malloc.

blakelead · Accepted Answer · 2015-10-14 18:16:33Z

0

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.

edited Oct 14, 2015 at 18:16

answered Oct 14, 2015 at 18:06

blakelead

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1 Comment

Rywi Over a year ago

I'm sorry for not clarifying one thing, what I want to do is remove is the * in the `*create*

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