Assign array to a different variable

Question

Here is a part of my code

int river(int a[], int n){
  int sorted[] = a;
}

n being the size of the array

I have tried a couple of different ways, including

sorted[n]

but I get problems each time. What I am trying to do is duplicate the array with a changed variable name. The array will be exactly the same, just stored in a different variable (because I want to obtain information but not modify the original array).

Is this just not possible because of something to do with dynamically changing the size of the array, or am I just missing the point?

Thanks!

That's what memcpy is for. " I want to obtain information but not modify the original array" -- wait ... then why do you want to copy it? Just use the original array. — Jim Balter
– Jim Balter, Commented Oct 6, 2014 at 7:14

juanchopanza · Accepted Answer · 2014-10-06 09:40:13Z

4

First of all, arrays are not assignable. You can either copy the contents of one array into another, or get a pointer to the first element of the array. For example

int a[42] = {0};
int* b = a;      // now you can access a's elements as b[0], b[1], etc.
int c[42];
memcpy(c, a, 42*sizeof(int)); //copy elements of a into c

Second, you should note that the function signature

int foo(int a[]);

is another way of saying

int foo(int* a);

i.e. it can take any pointer to int, even ones that don't point to arrays.

With that in mind, you can see that you easily lose array size information, which is why you need to pass the size as a parameter:

int foo(int a[], size_t n);

int a[42];
foo(a, 42);

Now, inside the function, you can copy the elements of one array into another:

int foo(int* a, size_t n)
{
  int sorted[n];
  memcpy(sorted, a, n * sizeof(int));
  // use sorted
  ....
  return 0;
}

edited Oct 6, 2014 at 9:40

answered Oct 6, 2014 at 7:05

juanchopanza

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2 Comments

alk Over a year ago

1+, but nitpicking: int n should be size_t n or even better const size_t n

juanchopanza Over a year ago

@alk Agreed. I was just basing it on OP's signature. But I'll change that.

timrau · Accepted Answer · 2014-10-06 07:02:08Z

1

You need to allocate space for the new copy of array by yourself.

int *sorted = malloc(n * sizeof(int));
memcpy(sorted, a, n * sizeof(int));

answered Oct 6, 2014 at 7:02

timrau

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2 Comments

mrybak834 Over a year ago

So it is a problem with memory allocation, interesting. Thank you very much!

Jim Balter Over a year ago

int sorted[n]; memcpy(sorted, a, n * sizeof *a); -- works too (if you're using a modern C compiler that supports C99 or later).

Anbu.Sankar · Accepted Answer · 2014-10-06 07:03:05Z

1

Simply try like this...

   for(int i=0;i<n;i++)
     sorted[i]=a[i];

If you want to use dynamic memory allocation malloc , try this

     int *sorted = malloc(n * sizeof(int));
     memcpy(sorted, a, n * sizeof(int));

answered Oct 6, 2014 at 7:03

Anbu.Sankar

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2 Comments

mrybak834 Over a year ago

Thank you for the answer, but I was trying more so to understand why my implementation wasn't working. Thanks for the help, however!

Jim Balter Over a year ago

Whether the array is dynamically allocated or not isn't relevant to whether to use memcpy or a loop ... both work in either case.

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