3 Easy Ways To Return Array From Function In C - 4

Table of Contents

Return array from function in C

Sometimes we want our function to return an array but in C programming it is not allowed to return the complete array.
There are 3 ways to solve this problem.
1. By declaring a local array variable as static.
2. Using dynamically allocated array
3. With help of structure

C program to return an array from function

By declaring a local array variable as static.

we can return a pointer to an array.
While returning a pointer we have to mention the array’s name without square brackets.
For this we have to create a function which returns a pointer.
Syntax for function which returns a pointer.
return_type * function_name(parameter_list)
{
body_of_function;
}

#include<stdio.h>

int * myfunction()
{
    int array1[] = {1,2,3,4,5};
    return array1;
}

int main()
{
    int *ptr,i;
    ptr = myfunction();
    for(i=0;i<5;i++)
    {
        printf("nptr[%d] = %d",i,ptr[i]);
    }
    return 0;
}

Output:

raff.c: In function ‘myfunction’:
raff.c:6:12: warning: function returns address of local variable [-Wreturn-local-addr]
6 | return array1;
| ^~~~~~

We can see that above program will generate a warning i.e. “function returns address of local variable [-Wreturn-local-addr]“

This is because we are returning the address of the local variable, this is not preferable because the local variable will get destroyed from a memory once the function call is over.
To solve this issue we need to declare our local array variable as a static.

Correct code to return array from function using static keyword is given below.

#include <stdio.h>

int * myfunction()
{
    static int array1[] = {1,2,3,4,5};
    return array1;
}

int main()
{
    int *ptr,i;
    ptr = myfunction();
    for(i=0;i<5;i++)
    {
        printf("nptr[%d] = %d",i,ptr[i]);
    }
    return 0;
}

Output:

ptr[0] = 1
ptr[1] = 2
ptr[2] = 3
ptr[3] = 4
ptr[4] = 5

Using dynamically allocated array

In this we have to use a pointer in parameter of function which will point to the array which will be passed in function as argument.
We can access the original array with help of a pointer.

#include <stdio.h> 
int * fun(int *array1) 
{ 
    array1[0] = 1; 
    array1[1] = 2; 
    array1[2] = 3; 
    array1[3] = 4; 

    return array1; 
} 

int main() 
{ 
    int ary[4],i; 
    int * ptr = fun(ary); 
    for(i=0;i<4;i++)
    {
        printf("nptr[%d] = %d", i, ptr[i]); 
    }
    return 0; 
}

Output:

ptr[0] = 1
ptr[1] = 2
ptr[2] = 3
ptr[3] = 4

With help of structure

In this we have to create a structure so that we can create/wrap an array in it.
Once the structure is created we have to create a function which returns an instance of that function.
We can access the array from the instance of a structure.

#include <stdio.h> 

struct struct1 
{ 
    int array1[10]; 
}; 

struct struct1 myfunction() 
{ 
    struct struct1 a; 

    a.array1[0] = 1; 
    a.array1[1] = 2; 
    a.array1[2] = 3; 
    a.array1[3] = 4; 
    a.array1[4] = 5;

    return a; 
} 

int main() 
{ 
    struct struct1 ary = myfunction(); 
    int i;
    for(i=0;i<5;i++)
    {
        printf("nary.array1[%d] = %d", i, ary.array1[i]); 
    }
    return 0; 
}