Passing array of structures to function by value in C

Question

I want to pass an array of three structures to a function that sorts the three numbers stored in the array of structures, but do not want to change the actual values , as arrays are passed by reference , there's my problem , any ideas accept thanks :)

Here is my code for review:

#include <stdio.h>
#define N 3
typedef struct
{
    int digito;
} numbers;

void  sort(numbers tabla[], int nro);
void enternums(numbers  tabla[], int nro);

int main()
{
    static numbers enter[N];
    int i, j, tmp;
    enternums(enter, N);
    printf("Data before entering the function...\n\n");
    for (i = 0 ; i < N ; i++)
    {
        printf("%d\n", enter[i].digito);
    }
    sort(enter, N);

    printf("After entering data function....\n\n");
    for (i = 0 ; i < N ; i++)
    {
        printf("%d\n", enter[i].digito);
    }
    return 0;
}

void enternums(numbers tabla[], int nro)
{
    int i;
    for (i = 0 ; i < nro ; i++)
    {
        printf("Enter a digit for the position %d:  ", i);
        scanf("%d", &tabla[i].digito);
    }
}

void sort(numbers tabla[N], int nro)
{
    int i, j, tmp;
    for (i = 1 ; i < nro ; i++)
    {
        j = i;
        tmp = tabla[i].digito;
        while ((j > 0) && (tmp < tabla[j - 1].digito))
        {
            tabla[j].digito = tabla[j - 1].digito;
            j--;
        }
        tabla[j].digito = tmp;
    }
}

You don't need to write the code in such a way that it's so hard to read, you can do it like I did. And certainly arrays are not passed by reference. — Iharob Al Asimi
– Iharob Al Asimi, Commented Oct 18, 2015 at 3:48
I don't really understand what you actually want, and scanf("%d", &tabla[i].digito); is a potential bug in your code since you don't check for the returned value, i.e. success or failure of scanf(). — Iharob Al Asimi
– Iharob Al Asimi, Commented Oct 18, 2015 at 3:54
@bpgeck Which would make sens if you return the array, otherwise it's just an intentional memory leak! — Iharob Al Asimi
– Iharob Al Asimi, Commented Oct 18, 2015 at 3:54
@iharob He doesn't have to do a deep copy. He can just store it on the stack. — bpgeck
– bpgeck, Commented Oct 18, 2015 at 3:55

Iharob Al Asimi · Accepted Answer · 2015-10-18 04:02:17Z

Passing the array as a pointer¹ to it's first element, has the advantage that you can actually pass another array to store the new values

void sort(const numbers *const tabla, numbers *resultado, int nro)
{
    int i, j, tmp;
    for (i = 1 ; i < nro ; i++)
        resultado[i].digito = tabla[i].digito;
    for (i = 1 ; i < nro ; i++)
    {
        tmp = resultado[i].digito;
        for (j = i ; (j > 0) && (tmp < resultado[j - 1].digito) ; --j)
            resultado[j].digito = resultado[j - 1].digito;
        resultado[j].digito = tmp;
    }
}

Making the while loop a for loop instead, it would make your code clearer.

And to use this

int main()
{
    static numbers enter[N];
    static numbers ordenado[N];
    int i, j, tmp;
    enternums(enter, N);
    printf("Data before entering the function...\n\n");
    for (i = 0 ; i < N ; i++)
    {
        printf("%d\n", enter[i].digito);
    }
    sort(enter, ordenado, N);

    printf("After entering data function....\n\n");
    for (i = 0 ; i < N ; i++)
    {
        printf("%d\n", ordenado[i].digito);
    }
    return 0;
}

Also, make your code safer by checking for possible errors, for example

scanf("%d", &tabla[i].digito);

should be

if (scanf("%d", &tabla[i].digito) != 1)
    Oops_there_is_a_problem_dont_use_this_value_maybe_ask_again();

¹_{Which happens automatically, it's not passed by reference it's converted to a pointer}.

R Sahu · Accepted Answer · 2015-10-18 03:49:46Z

0

You cannot pass arrays by value unless they are contained in a struct.

You can use:

typedef struct MyStrutArray
{
   numbers  enter[N];
} MyStructArray;

and then use:

void  sort(MyStructArray array);

answered Oct 18, 2015 at 3:49

R Sahu

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