Why does this Code use "call by reference"?

Question

In the below code, there is a problem that I didn't understand. In the lines of these:

gcd = GCD(n, A); 
lcm = LCM(n, A);

Despite I didn't use the pointers, the array A[n] elements has changed after the GCD function. Thus, the second function didn't work correctly. So, if I reversed the order of the above two lines, this time the GCD function wouldn't work correctly. What is my mistake?

/*
    Finding the greatest common divisor and the least common multiple in an array.
*/

#include <stdio.h>

int GCD(int n, int A[]);
int LCM(int n, int A[]);
int main()
{
    int n, gcd, lcm;
    printf("Length of the array?: ");
    scanf("%d", &n);
    int A[n];

    for (int i = 0; i < n; i++)
    {
        printf("Element #%d: ", i+1);
        scanf("%d", &A[i]);
    }

    gcd = GCD(n, A); 
    lcm = LCM(n, A);

    printf("\ngcd: %d\nlcm: %d\n", gcd, lcm);

    return 0;
}

int GCD(int n, int A[]) // Greatest common divisor
{
    int gcd = 1, j = 1, flag, ones = 0;

    while (1)
    {
        flag = 0;

        for (int i = 0; i < n; i++)
        {
            if (A[i] % j == 0)
            {
                A[i] /= j;
                flag++;

                if (A[i] == 1)
                    ones++;
            }
        }

        if (flag == 0 || j == 1)
            j++;
        else
            gcd *= j;

        if (ones == n)
            return gcd;
    }
}

int LCM(int n, int A[]) // Least common multiple
{
    int lcm = 1, j = 2, flag = 0, ones = 0;

    for (int i = 0; i < n; i++)
        if (A[i] == 1)
            flag++;

    if (flag != 0)
        return 1;

    while (1)
    {
        for (int i = 0; i < n; i++)
        {
            if (A[i] % j == 0)
            {
                A[i] /= j;
                flag++;

                if (A[i] == 1)
                    ones++;
            }
        }

        if (flag == n)
            lcm *= j;

        if (flag == 0)
            j++;

        if (ones == n)
            return lcm;

        flag = 0;
    }
}

`int A[]` in a parameter list means `int *A`. You're using pointers. — melpomene, Apr 29 '17 at 20:55

score 2 · Accepted Answer · edited May 23 '17 at 11:54

2

Arrays in C are converted, in most of the cases, to a pointer to the first element of the array itself. And more in detail arrays passed into functions are always converted into pointers.

Here a quote from K&R2nd:

When an array name is passed to a function, what is passed is the location of the initial element. Within the called function, this argument is a local variable, and so an array name parameter is a pointer, that is, a variable containing an address.

Writing:

int GCD(int n, int A[]);

has the same meaning as writing:

int GCD(int n, int * A);

So despite you are not writing it explicitly it is as you are passing a pointer.

For more I really suggest reading this.

Moreover, you can find other answers on SO here and here

edited May 23 '17 at 11:54

Community

1
1

answered Apr 29 '17 at 20:56

granmirupa

2,780
16
27

2

Any explanation that involves "the name of" isn't right. It's like saying `int i = 42; double d = i;` works because "the name of an integer is converted to a floating point number". – melpomene Apr 29 '17 at 20:57
It is an explanation, you just have to remove "name of". – melpomene Apr 29 '17 at 21:04
@melpomene ok thanks for your tip – granmirupa Apr 29 '17 at 21:06
Thank you! So, is only way to fix this, am i need to create a new array and copy elements of A[n] to the new array? (Then i can send two same arrays to the two different functions.) – Atreidex Apr 29 '17 at 21:13
@LionKlan if you need to modify the array into the function, yes – granmirupa Apr 29 '17 at 21:23
it's not "converted", but *decayed* – phuclv Apr 30 '17 at 02:09
1

@LưuVĩnhPhúc the standard does not use the term "decay". It uses "converted" (for expressions) and "adjusted" (for parameters). OP's first sentence is mostly correct, although a reader could get a false impression that it is an explanation for the rest of the answer – M.M Apr 30 '17 at 03:09
@M.M what do you suggest in order to improve my answer? – granmirupa Apr 30 '17 at 09:11
@granmirupa doesn't really matter as there are many duplicates. Maybe remove the first sentence though or clarify that adjustment and conversion are different – M.M Apr 30 '17 at 09:49
@M.M maybe now it is a better answer that explain this specific case without too much divagation. – granmirupa Apr 30 '17 at 10:46

Why does this Code use "call by reference"?

1 Answers1