How to pass an array of structures to a function by reference in C?

Question

This code is similar to what I am attempting to do, however I am getting errors saying that I am passing incompatable types

 #include <stdio.h>

struct numbers{
    int num;
};

void fillArray(struct numbers* a[]);

int main(void)
{
    struct numbers array[4]; 

    fillArray(&array);

    for(int i = 0; i < 4; i++)
    {
        printf("%d", array[i].num);
    }

}

void fillArray(struct numbers* a[])
{
    for(int i = 0; i < 4; i++)
    {
        a[i]->num = i;
    }
}

You need to include the error message(s) in your question. – Kninnug Apr 10 '17 at 09:46 — Kninnug, Apr 10 '17 at 09:46

Vlad from Moscow · Answer 1 · 2017-04-10T10:23:38.747

The function parameter

void fillArray(struct numbers* a[]);

is adjusted to

void fillArray(struct numbers ** a);

On the other hand the type of the argument in this call

fillArray(&array);

is struct numbers( * )[4]. The types struct numbers ** and struct numbers ( * )[4] are incompatible.

There is no need to pass a pointer to the array because elements of the array are already passed indirectly if you will pass just the array that is implicitly converted to pointer to its first element.

So what you need is to declare and define the function like

void fillArray( struct numbers a[] )
// or
// void fillArray( struct numbers *a )
{
    for(int i = 0; i < 4; i++)
    {
        a[i].num = i;
    }
}

and call it like

fillArray( array );

Take into account that the function depends on magic number 4. It is better to define the function such a way that it could deal with arrays of various numbers of elements.

So I would define the function like

void fillArray( struct numbers a[], size_t n )
// or
// void fillArray( struct numbers *a, size_t n )
{
    for ( size_t i = 0; i < n; i++ )
    {
        a[i].num = i;
    }
}

and call it like

fillArray( array, 4 );

Here is demonstrated how the program can look in whole

#include <stdio.h>

struct numbers
{
    int num;
};

void fillArray( struct numbers a[], size_t n )
{
    for ( size_t i = 0; i < n; i++ )
    {
        a[i].num = i;
    }
}

#define N   4

int main(void) 
{
    struct numbers array[N]; 

    fillArray( array, N );

    for ( size_t i = 0; i < N; i++ )
    {
        printf( "%d ", array[i].num );
    }

    putchar( '\n' );

    return 0;
}

Its output is

0 1 2 3

To use an array with a different number of elements it is enough to change the value of the macro name N. Thus the program and its function do not depend on the magic number 4.

score 1 · Answer 2 · edited May 23 '17 at 12:02

Your function wants an array of pointers, but you have a single array, so

void fillArray(struct numbers* a);

or

void fillArray(struct numbers a[]);

Moreover using a[i] you are already dereferencing the pointer, so you need . not -> operator.

#include <stdio.h>

struct numbers
{
    int num;
};

void fillArray(struct numbers *a);

int main(void)
{
    struct numbers array[4];

    fillArray(array);

    for (int i = 0; i < 4; i++)
    {
        printf("%d", array[i].num);
    }

    printf("\n");

}

void fillArray(struct numbers *a)
{
    for (int i = 0; i < 4; i++)
    {
        a[i].num = i;
    }
}

Finally using c array decays to pointer to first item of array, so

fillArray(&array);

must be

fillArray(array);

or

fillArray(&array[0]);

Lastly you should pass size of array to your function, instead of using fixed numbers. You can do it using pointer and size

#include <stdio.h>

struct numbers
{
    int num;
};

void fillArray(size_t size, struct numbers *a);

int main(void)
{
    struct numbers array[4];

    fillArray(sizeof(array)/sizeof(array[0]), array);

    for (int i = 0; i < 4; i++)
    {
        printf("%d", array[i].num);
    }

    printf("\n");

}

void fillArray(size_t size, struct numbers *a)
{
    for (size_t i = 0; i < size; i++)
    {
        a[i].num = i;
    }
}

Or using VLAs

#include <stdio.h>

struct numbers
{
    int num;
};

void fillArray(size_t size, struct numbers a[size]);

int main(void)
{
    struct numbers array[4];

    fillArray(sizeof(array)/sizeof(array[0]), array);

    for (int i = 0; i < 4; i++)
    {
        printf("%d", array[i].num);
    }

    printf("\n");

}

void fillArray(size_t size, struct numbers a[size])
{
    for (size_t i = 0; i < size; i++)
    {
        a[i].num = i;
    }
}

`fillArray(&array);` : But compiler say that _incompatible pointer type_ — BLUEPIXY, Apr 10 '17 at 10:00
@BLUEPIXY Yes,I edited.... while `printf("%p %p\n", (void *)(&array), (void *)array);` prints the same address... ;) — LPs, Apr 10 '17 at 10:13

score 1 · Answer 3 · answered Apr 10 '17 at 09:46

1

You probably want this:

#include <stdio.h>

struct numbers {
    int num;
};

void fillArray(struct numbers a[]);

int main(void)
{
    struct numbers array[4];

    fillArray(array);

    for (int i = 0; i < 4; i++)
    {
        printf("%d\n", array[i].num);
    }

}

void fillArray(struct numbers a[])
{
    for (int i = 0; i < 4; i++)
    {
        a[i].num = i;
    }
}

answered Apr 10 '17 at 09:46

Jabberwocky

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Thank you, does this method work because an array is a pointer to the first element in C? – V. Winters Apr 10 '17 at 09:51
Yes, I tried it and it works perfectly. I needed tpo know it for a simple electronic boardgame I'm coding in C. I wasn't sure how to pass player data to one of the functions I'm using – V. Winters Apr 10 '17 at 09:56
@V.Winters sorry, I misundertstood your comment: yes, the array name is actually more or less the same than the pointer to the first element, but you can't modify it. E.g: `int *a = ...; a++;` is OK, but `int a[5]; ... a++;` is not. – Jabberwocky Apr 10 '17 at 09:59
@V.Winters If the answer is good for you you should close the question as answered selecting this answer. – LPs Apr 10 '17 at 11:53
@V.Winters C is a bit intricate and weird here. When you pass the array to the function, it does indeed decay to a pointer to the first element. As for the function itself, the parameter `struct numbers a[]` also decays into a pointer to the first element. So you end up passing a `struct numbers*` pointer to a function accepting a `struct numbers*`, which is why it works. This design is intentional by C, to block the programmer from accidentally passing the whole array by value to the function, which would have been very inefficient. – Lundin Apr 10 '17 at 12:56

score -1 · Answer 4 · answered Apr 10 '17 at 09:43

It is sufficient to only mention the array name. The compiler will transform it into a pointer to the first element.

So instead of

 fillArray(&array);

you write

 fillArray(array);

Though this answers your question, of course the function being called must have a compatible definition, which in your case is not so.

How to pass an array of structures to a function by reference in C?

4 Answers4