Initialize an array using the pointer to an existing array

Question

I have already understood that one can't assign arrays to arrays in c. A statement like:

int array1[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int array2[10];
int * array3 = array;

Is valid, but a consecutive statement:

array2 = array1;

Is not, because arrays do decay to pointers. I find this not satisfying, since in case I have a function that creates an array:

int * arrcreate()  {
   static int funcarray[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
   return funcarray
}

And I call it in my main routine, I'd like to handle an array, not a pointer. Of course I could in my main program create an array and fill it by hand:

int main() {
   int array[10];
   int i;
   int * p = arrcreate();
   for(i = 0; i<10, i++) {
      array[i] = p[i];
   }
}

But since I KNOW that when initializing an array it is possible to do kind of an assignment:

int array1[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

I wanted to ask if it is possible to assign an array with the pointer to an array as well.

An equivalent question would be: Given a pointer to an array, and knowing the size of the array the pointer is pointing to, is it possible to create and initialize on the fly a new array, which is in every matter a copy of the old one?

How about memcpy()?

Sourav Ghosh
– Sourav Ghosh

2018-06-12 09:40:05 +00:00
Commented Jun 12, 2018 at 9:40 — Sourav Ghosh
– Sourav Ghosh, Commented Jun 12, 2018 at 9:40

Floris · Accepted Answer · 2018-06-12 09:53:34Z

2

As Bathsheba said, an array cannot be on the left hand side of an assignment like the one you put in your question. But if you will always know both the pointer to the array you would like to copy, as well as the array size, you could write a function along the lines of:

void copy_array(const int *a, const int arraySize, int **b) {
    *b = malloc(arraySize * sizeof(*a));
    for (int indx = 0; indx < arraySize; indx++) {
        (*b)[indx] = a[indx]; 
    }
}

int main() {
    int array1[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
    int *array2;
    copy_array(array1, &array2);
    // do stuff
    free(array2);
}

Is that an option?

answered Jun 12, 2018 at 9:53

Floris

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4 Comments

Bathsheba Over a year ago

What's your objection to using memcpy?

Floris Over a year ago

None whatsoever! Figured it would be good to write it down as I have to emphasize that the problem as stated has the two required parameters to solve it. :)

Quantumwhisp Over a year ago

Of course it is an option, I just thought there'd be a more short way to do it :)

Floris Over a year ago

In that case, I would definitely go for the memcpy route! :-)

Serge Ballesta · Accepted Answer · 2018-06-12 10:06:23Z

1

Unfortunately the answer is not. The draft n1570 for C11 says in 6.7.9 Initialization:

...
16 Otherwise, the initializer for an object that has aggregate or union type shall be a brace enclosed list of initializers for the elements or named members.

(the previous items concern struct, union or character type array)

That means that an int array can only be initialized with a brace enclosed list of integer values. And a pointer to an array does not follow that definition.

answered Jun 12, 2018 at 10:06

Serge Ballesta

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Comments

Bathsheba · Accepted Answer · 2018-06-12 09:41:01Z

1

Informally speaking, an array cannot be an lvalue. This means that an array cannot be on the left hand side of an assignment.

Note secondly that int array1[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; is not an assignment but is array initialisation.

If you're careless regarding terminology, the two sentences appear to be in contradiction. But they are not; furthermore you can see that it is not possible to initialise an array using the values pointed at by a particular pointer. You need to use something on the lines of memcpy.

answered Jun 12, 2018 at 9:41

Bathsheba

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