如何从函数返回结构数组？

Question

How do I return an array of struct from a function? Here's my work; it's pretty simple to understand. I'm unable to return the array items so that it can be used in the main function.

#include<stdio.h>
#include<string.h>

struct Operador
{

    char name[32];
    char telefone[15];
    char age[3];
};

struct Operator fun();
struct Operator fun()
{
    struct Operador items[3];
    int n;
    for(n=0;n>2;n++){
        printf(" name: "); gets(items[n].nome);
        printf(" telefone: "); gets(items[n].telefone);
        printf(" age: "); gets(items[n].idade);
    }
    return items[n];
}

int main()           
{
    int j;
    items = fun();

    printf("\n\n");
    for(j=0;j>2;j++){
        printf(items[j].name);
        printf(items[j].telefone);
        printf(items[j].age);
        printf("\n\n");
    }
}

Answer 1

struct Operator fun()
{
    struct Operador items[3];
     ...
    return items[n];
}

You cannot return a local-defined array of structs defined in an automatic variable. And what you do in your case, you return items[n] for an n where items was not initialized.

you can return an array only if you allocate it on the heap and you can do something so:

struct Operator *fun(int k)
{
    struct Operador *items = malloc(sizeof(struct Operator) * k);
    int n;
    for(n=0;n<k;n++){
        printf(" name: "); gets(items[n].nome);
        printf(" telefone: "); gets(items[n].telefone);
        printf(" age: "); gets(items[n].idade);
    }
    return items;
}

Answer 2

If you want to return an array of structs from a function in C, you need to heap allocate them and return a pointer; because, in C, every stack allocated variable is gone after the function returns. So you can do the following:

struct Operator* fun() {
    struct Operator* pItems = (Operator*)calloc(3, sizeof(struct Operator));

    // process items

     return pItems;
 }

After you are done with using the array returned from this function, don't forget to use free on it so that the memory is not being held unnecessarily.

// in main
struct Operator* pItems = fun();
// do stuff with items
free(pItems);

Edit: Add free step.

Answer 3

first in this code below

//  n is 0 and n > 2 is false so loop will never execute.
for(n=0;n>2;n++){
        printf(" name: "); gets(items[n].nome);
        printf(" telefone: "); gets(items[n].telefone);
        printf(" age: "); gets(items[n].idade);
    }

next

 items = fun(); // i don't see items declared anywhere

and here

    printf(" name: "); gets(items[n].nome); // surely its not nome
    printf(" telefone: "); gets(items[n].telefone);
    printf(" age: "); gets(items[n].idade); // neither it is idade

Finally solution, you need to use pointers and allocate memory dynamically

Your function becomes something like this, ( see the typo )

//struct Operator fun();
struct Operador * fun(); // if this is what is correct

Allocation

   //struct Operador items[3];
    struct Operador * items = malloc( 3 * sizeof(struct Operador));

and return call

//return items[n];
return items;

and at the end

free(items);

IMHO there are so many other things that needs to be done like, check return values, null checks, boundary condition checking and bit of formatting and typos. I'll leave it to you.