CODEWITHSUNDEEP
c
seasick
seasick

Reputation: 1225

Return integer array from function

I'm trying to return an array of integers from a function, sort the numbers then pass everything back to main. I haven't allocated and freed memory in this piece of code. I was just trying to see if it would actually work. The compiler flags an error for the statement b=sort(a). It says that it is not assignable, which would make sense. The input integers are not pointers. Is there a way to declare an array of integers as pointers? such as :

int *a[5]={3,4} 

#include <stdio.h>
#include <stdlib.h>
int *sort(int *input_array);

int *sort(int *input_array)
{
    return input_array;
}

int main()
{
    int a[5]={3,4};
    int b[5];
    b=sort(a);
    return 0;
}

Upvotes: 8

Views: 42217

Answers (4)

aaaaaa123456789
aaaaaa123456789

Reputation: 5842

When you create an array, you cannot assign to the array itself (only to the elements). Besides, since when you pass an array, you're passing it by reference, sort() would modify the array, making it unneeded to return it.

What you're looking for is either of: sorting the original array, which would be like this:

void sort (int * array);

void sort (int * array) {
  // do stuff on the array
}

int main (void) {
  int a[5] = {1, 46, 52, -2, 33};
  sort(a); // result is still in a
  return 0;
}

Or creating a copy and sorting it, which would be like this:

#include <stdlib.h>
#include <string.h>
int * sort (int * array, unsigned size);

int * sort (int * array, unsigned size) {
  int * copy = malloc(sizeof(int) * size);
  memcpy(copy, array, size * sizeof(int));
  // sort it somehow
  return copy;
}

int main (void) {
  int a[5] = {1, 46, 52, -2, 33};
  int * b; // pointer because I need to assign to the pointer itself
  b = sort(a, (sizeof a) / (sizeof *a)); // now result is in b, a is unchanged
  // do something with b
  free(b); // you have to
  return 0;
}

Upvotes: 12

unwind
unwind

Reputation: 399703

You can't assign arrays, they're not "first class citizens" but instead behave much like pointers.

You need something like:

int a[] = { 3, 4 };
int *b;

b = sort(a, sizeof a / sizeof *a);

The sizeof expression is needed to compute the length of the array, the sort() function can't determine that from the bare pointer it gets passed.

UPDATE: The above assumes that you won't be changing the input array, but if you do then (as pointed out in a comment, thanks) the return value is of course not needed since the caller's a will have changed when the sort() call returns.

Upvotes: 5

xaxxon
xaxxon

Reputation: 19751

You can't return an array of anything in C. You can only return a single instance of a single datatype.

That datatype can be a pointer to memory storing a sequential list of numbers (or anything else), but you lose all information about how long the result is, so you either need to know that, or you have to have another value as an output variable to tell you the length.

You can also return a custom datatype, such as a struct, that would contain both the list of data as well as the length. However, returning a large datastructure creates multiple shallow copies of the data structure, slowing down execution of your program as well as creating memory nightmares with leaks and multiple references.

Returning a pointer to a custom data structure, however, can work very well.

Upvotes: 1

Rohan
Rohan

Reputation: 53316

If you are passing array - a pointer of int, you don't need to return a changed array. The array that you passed will get changed.

As @unwind suggested, you should pass number of elements to the function also so that the function knows how many elements are there in the array.

Upvotes: 1

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