CODEWITHSUNDEEP
carraysreturn
Kostas
Kostas

Reputation: 1369

Return an array in c

I would like to know if there is any way to return an char array. I tried something like this "char[] fun()" but I am getting error.

I don't want a pointer solution. Thanks!

Upvotes: 4

Views: 5051

Answers (5)

Ruchir
Ruchir

Reputation: 1068

A Very very basic code and very very basic explanation HOW TO Return array back from a user defined function to main function..Hope It helps!! Below I have given complete code to make any one understand how exactly it works? :) :)

#include<iostream>
using namespace std;

char * function_Random()
{
    int i;
    char arr[2];
    char j=65;//an ascII value 65=A and 66=B
    cout<<"We are Inside FunctionRandom"<<endl;
    for(i=0;i<2;i++)
    {
        arr[i]=j++;// first arr[0]=65=A and then 66=B
        cout<<"\t"<<arr[i];
    }
    cout<<endl<<endl;
    return arr;
}
int main()
{

    char *arrptr;
    arrptr=function_Random();
    cout<<"We are Inside Main"<<endl;
    for(int j=0;j<2;j++)
    {
        cout<<"\t"<<arrptr[j];
    }
    return 0;
}

Upvotes: -1

anon
anon

Reputation:

You can return an array by wrapping it in a struct:

struct S {
   char a[100];
};

struct S f() {
    struct S s;
    strcpy( s.a, "foobar" );
    return s;
}

Upvotes: 11

John Bode
John Bode

Reputation: 123596

C functions cannot return array types. Function return types can be anything other than "array of T" or "function returning T". Note also that you cannot assign array types; i.e., code like the following won't work:

int a[10];
a = foo();

Arrays in C are treated differently than other types; in most contexts, the type of an array expression is implicitly converted ("decays") from "N-element array of T" to "pointer to T", and its value is set to point to the first element in the array. The exceptions to this rule are when the array expression is an operand of either the sizeof or address-of (&) operators, or when the expression is a string literal being used to initialize another array in a declaration.

Given the declaration 

T a[N];

for any type T, then the following are true:

Expression     Type      Decays to      Notes
----------     ----      ---------      -----
         a     T [N]     T *            Value is address of first element
        &a     T (*)[N]  n/a            Value is address of array (which
                                          is the same as the address of the
                                          first element, but the types are
                                          different)
  sizeof a     size_t    n/a            Number of bytes (chars) in array = 
                                          N * sizeof(T)
sizeof a[i]    size_t    n/a            Number of bytes in single element = 
                                          sizeof(T)
       a[i]    T         n/a            Value of i'th element
      &a[i]    T *       n/a            Address of i'th element

Because of the implicit conversion rule, when you pass an array argument to a function, what the function receives is a pointer value, not an array value:

int a[10];
...
foo(a);
...

void foo(int *a)
{
  // do something with a
}

Note also that doing something like

int *foo(void)
{
  int arr[N];
  ...
  return arr;
}

doesn't work; one the function exits, the array arr technically no longer exists, and its contents may be overwritten before you get a chance to use it.

If you are not dynamically allocating buffers, your best bet is to pass the arrays you want to modify as arguments to the function, along with their size (since the function only receives a pointer value, it cannot tell how big the array is):

int a[10];
init(a, sizeof a / sizeof a[0]);  // divide the total number of bytes in 
...                               // in the array by the number of bytes
void init(int *a, size_t len)     // a single element to get the number
{                                 // of elements
  size_t i;
  for (i = 0; i < len; i++)
    a[i] = i;
}

Upvotes: 4

Roger Pate
Roger Pate

Reputation:

Arrays cannot be passed or returned by value in C.

You will need to either accept a pointer and a size for a buffer to store your results, or you will have to return a different type, such as a pointer. The former is often preferred, but doesn't always fit.

Upvotes: 5

jk.
jk.

Reputation: 14004

arrays aren't 1st class objects in C, you have to deal with them via pointers, if the array is created in your function you will also have to ensure its on the heap and the caller cleans up the memory

Upvotes: 3

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