passing back values from array in struct by ref - c

Question

struct myStruct
{
int* arr;
int size;
};
void get_back(struct myStruct* my ,int* arr, int* size)
{
arr = my->arr;
*size = my->size;
}
int main()
{
struct myStruct my;
my.arr = (int*) malloc(3 * sizeof(int));
my.arr[0] = 20;
my.arr[1] = 200;
my.arr[2] = 2000;
my.size = 3;
int* ret_arr = NULL;
int size;
get_back(&my, ret_arr, &size);
free(my.arr);
return 1;
}
The goal of my simple program is to get back the values from my.arr into ret_arr, since ret_arr=nullptr, do I need to allocate the memory and than copy it into the array inside get_back function?
or I can just point to the existing array inside "my" struct?
This is my current solution, I copy the values.
struct myStruct
{
int* arr;
int size;
};
int* get_back(struct myStruct* my , int* size)
{
int *arr = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < my->size; i++)
{
arr[i] = my->arr[i];
}
*size = my->size;
return arr;
}
int main()
{
myStruct my;
my.arr = (int*) malloc(3 * sizeof(int));
my.arr[0] = 20;
my.arr[1] = 200;
my.arr[2] = 2000;
my.size = 3;
int* ret_arr = NULL;
int size;
ret_arr = get_back(&my, &size);
free(my.arr);
free(ret_arr);
return 1;
}

Answer 1

Seeing as you're freeing the array, you probably want to copy the contents over with memcpy (from string.h).
You will also need to include stdlib.h for malloc.
#include <string.h>
#include <stdlib.h>
// Struct definition goes here
void get_back(struct myStruct* my, int** arr, int* size)
{
*arr=malloc(my->size*sizeof(int)); //Allocate space for integers
memcpy(*arr, my->arr, my->size*sizeof(int)); //Copy integers to new array
*size=my->size;
}
The function needs to take a pointer to the pointer in order to be able to modify it.
Additionally, your main function will need to be modified too.
int main()
{
struct myStruct my; // Structs are not types.
my.arr = (int*) malloc(3 * sizeof(int));
my.arr[0] = 20;
my.arr[1] = 200;
my.arr[2] = 2000;
my.size = 3;
int* ret_arr = NULL;
int size;
get_back(&my, &ret_arr, &size); //Need to pass ret_arr by reference
free(my.arr);
return 1;
}

Answer 2

Use std::vector, it's very comfortable and has many useful algorithms, there is std::copy function to copy from one vector to another, take a look at your task with a help of vectors:
#include <vector>
#include <iostream>
struct myStruct
{
std::vector<int> arr;
};
int main()
{
myStruct my;
my.arr.push_back(20);
my.arr.push_back(200);
my.arr.push_back(2000);
std::vector<int> ret_arr;
std::copy(my.arr.begin(), my.arr.end(), std::back_inserter(ret_arr));
return 1;
}
And the result is on screen:
If you want to use C language then you should pass pointer on pointer(int** arr) to get right pointing after leaving scope. I will show two methods, first just to point on already allocated memory:
void get_back_pointers(myStruct* my ,int** arr, int* size)
{
*arr = my->arr;
*size = my->size;
}
Other is for deep copy, to allocate new array and copy data to it:
void get_back_copy(myStruct* my ,int** arr, int& size)
{
*arr = (int*) malloc(3 * sizeof(int));
memcpy( *arr, my->arr, my->size * sizeof(int) );
size = my->size;
}
After get_back_copy passed arr will be needed to free its memory. In debugger you can see results that my.arr and ret_arr have one address but ret_arr2 has another because it's allocated in new memory: